Tanuja Puram

Integration Using Partial Fractions

Partial Fractions

Rational Functions If f(x) and g(x) are polynomial functions such that g(x) ≠ 0 then \(\frac{f(x)}{g(x)}\) is called a rational function.

If degree f(x) < degree g(x) then \(\frac{f(x)}{g(x)}\) is called a proper rational function.

If degree f(x) ≥ degree then \(\frac{f(x)}{g(x)}\) is called an improper rational function.

If \(\frac{f(x)}{g(x)}\) is an improper rational function then by dividing f(x) by g(x), we can express \(\frac{f(x)}{g(x)}\) as the sum of a polynomial and a proper rational function.

Partial Fractions Any proper rational function \(\frac{p(x)}{q(x)}\) can be expressed as the sum of rational functions, each having a simplest factor q(x). Each such fraction is known as a partial fraction and the process of obtaining them is called the decomposition or resolving of the given function into partial fractions.

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Integration By Partial Fractions Formula

Method We first resolve the denominator of the given fraction into simplest factors. Based on these factors, we obtain the corresponding partial fraction as per rules given below:

The values of A, B, C, etc., can be obtained as shown below.

Solved Examples On Partial Fractions

Example 1 Resolve \(\frac{2 x+3}{(x-3)(x+1)}\) into partial fractions.

Solution

Given

\(\frac{2 x+3}{(x-3)(x+1)}\)

Let \(\frac{(2 x+3)}{(x-3)(x+1)}=\frac{A}{(x-3)}+\frac{B}{(x+1)} .\)

\(\frac{2 x+3}{(x-3)(x+1)}=\frac{A(x+1)+B(x-3)}{(x-3)(x+1)}\)

or (2x + 3) ≡ A(x+1) + B(x-3) …(1)

Putting (x-3) = 0 or x = 3 in (1), we get A = (9/4).

Putting (x+1) = 0 or x = -1 in (1), we get B = (-1/4).

∴ \(\frac{(2 x+3)}{(x-3)(x+1)}=\frac{9}{4(x-3)}-\frac{1}{4(x+1)}\)

Integration By Partial Fractions Formula

Example 2 Resolve \(\frac{x^3-2 x^2-13 x-12}{x^2-3 x-10}\) into partial fractions.

Solution

Given

\(\frac{x^3-2 x^2-13 x-12}{x^2-3 x-10}\)

On dividing, we get

\(\frac{x^3-2 x^2-13 x-12}{x^2-3 x-10}=(x+1)-\frac{2}{\left(x^2-3 x-10\right)}\) …(1)

Let \(\frac{2}{\left(x^2-3 x-10\right)}=\frac{2}{(x-5)(x+2)}=\frac{A}{x-5}+\frac{B}{x+2}\)

Then, \(\frac{2}{(x-5)(x+2)}=\frac{A(x+2)+B(x-5)}{(x-5)(x+2)}\)

or 2 ≡ A(x+2) + B(x-5) …(2)

Putting (x-5) = 0 or x = 5 in (2), we get A = (2/7).

Putting (x+2) = 0 or x = -2 in (2), we get B = (-2/7).

∴ \(\frac{2}{\left(x^2-3 x-10\right)}=\frac{2}{7(x-5)}-\frac{2}{7(x+2)}\)

Hence, \(\frac{x^3-2 x^2-13 x-12}{x^2-3 x-10}=(x+1)-\frac{2}{7(x-5)}+\frac{2}{7(x+2)}\).

Understanding Integration by Partial Fractions

Example 3 Resolve \(\frac{16}{(x-2)(x+2)^2}\)

Solution

Given

\(\frac{16}{(x-2)(x+2)^2}\)

Let \(\frac{16}{(x-2)(x+2)^2}=\frac{A}{x-2}+\frac{B}{x+2}+\frac{C}{(x+2)^2}\)

or \(\frac{16}{(x-2)(x+2)^2}=\frac{A(x+2)^2+B(x-2)(x+2)+C(x-2)}{(x-2)(x+2)^2}\)

∴ 16 ≡ A(x+2)² + B(x-2)(x+2) + C(x-2) …(1)

or 16 ≡ (a+b)x² + (4A+C)x + (4A-4B-2C) …(2)

Putting (x-2) = 0 or x = 2 in (1), we get A = 1.

Putting (x+2) = 0 or x = -2 in (1), we get C = -4.

Comparing coefficients of x² on both sides of (2), we get

A + B = 0 or B = -A = -1.

Thus A = 1, B = -1 and C = -4.

∴ \(\frac{16}{(x-2)(x+2)^2}=\left[\frac{1}{(x-2)}-\frac{1}{(x+2)}-\frac{4}{(x+2)^2}\right]\)

Different Forms of Partial Fraction Decomposition

Example 4 Resolve \(\frac{2 x+1}{(x-1)\left(x^2+1\right)}\) into partial fractions.

Solution

Given

\(\frac{2 x+1}{(x-1)\left(x^2+1\right)}\)

Let \(\frac{2 x+1}{(x-1)\left(x^2+1\right)}=\frac{A}{(x-1)}+\frac{B x+C}{\left(x^2+1\right)}\)

or \(\frac{(2 x+1)}{(x-1)\left(x^2+1\right)}=\frac{A\left(x^2+1\right)+(B x+C)(x-1)}{(x-1)\left(x^2+1\right)} .\)

∴ 2x + 1 ≡ A(x²+1) + (Bx+C)(x-1)

or 2x + 1 ≡ (A+B)x² + (C-B)x + (A-c)  …(1)

Equating the like powers of x on both sides of (1), we get

A + B = 0, C – B = 2 and A – C = 1.

On solving these equations, we get

A = \(\frac{3}{2}\), B = \(-\frac{3}{2}\) and C = \(\frac{1}{2}\).

∴ \(\frac{2 x+1}{(x-1)\left(x^2+1\right)}=\frac{3}{2(x-1)}+\frac{\left(\frac{-3}{2} x+\frac{1}{2}\right)}{x^2+1}=\left[\frac{3}{2(x-1)}+\frac{(1-3 x)}{2\left(x^2+1\right)}\right] .\)

Integration Using Partial Fractions

Integration Using Partial Fractions Solved Examples

Example 1 Evaluate \(\int \frac{(x-1)}{(x+1)(x-2)} d x\).

Solution

Given:

\(\int \frac{(x-1)}{(x+1)(x-2)} d x\)

Let \(\frac{(x-1)}{(x+1)(x-2)}=\frac{A}{(x+1)}+\frac{B}{(x-2)} .\) …(1)

Then, (x-1) ≡ A(x-2) + B(x+1)

Putting x = -1 in (1), we get A = \(\frac{2}{3}\).

Putting x = 2 in (1), we get B = \(\frac{1}{3}\).

∴ \(\frac{(x-1)}{(x+1)(x-2)}=\frac{2}{3(x+1)}+\frac{1}{3(x-2)}\)

⇒ \(\int \frac{(x-1)}{(x+1)(x-2)} d x=\frac{2}{3} \int \frac{d x}{(x+1)}+\frac{1}{3} \int \frac{d x}{(x-2)}\)

= \(\frac{2}{3} \log |x+1|+\frac{1}{3} \log |x-2|+C \text {. }\)

\(\int \frac{(x-1)}{(x+1)(x-2)} d x\) = \(\frac{2}{3} \log |x+1|+\frac{1}{3} \log |x-2|+C \text {. }\)

Examples of Partial Fraction Integration

Example 2 Evaluate \(\int \frac{\left(x^2+1\right)}{\left(x^2-5 x+6\right)} d x\).

Solution

Given

\(\int \frac{\left(x^2+1\right)}{\left(x^2-5 x+6\right)} d x\)

Here the integrand is not a proper rational function on dividing (x²+1) by (x²-5x+6), we get

\(\frac{\left(x^2+1\right)}{\left(x^2-5 x+6\right)}=1+\frac{(5 x-5)}{\left(x^2-5 x+6\right)}=1+\frac{(5 x-5)}{(x-2)(x-3)} .\)

Now, let \(\frac{(5 x-5)}{(x-2)(x-3)}=\frac{A}{(x-2)}+\frac{B}{(x-3)}\)

⇒ \(\frac{(5 x-5)}{(x-2)(x-3)}=\frac{A(x-3)+B(x-2)}{(x-2)(x-3)}\)

⇒ (5x-5) ≡ A(x-3) + B(x-2) …(1)

Putting x = 2 on both sides of (1), we get A = -5.

Putting x = 3 on both sides of (1), we get B = 10.

∴ \(\frac{\left(x^2+1\right)}{\left(x^2-5 x+6\right)}=1-\frac{5}{(x-2)}+\frac{10}{(x-3)}\)

⇒ \(\int \frac{\left(x^2+1\right)}{\left(x^2-5 x+6\right)} d x=\int d x-5 \int \frac{d x}{(x-2)}+10 \int \frac{d x}{(x-3)}\)

= X = 5log |x-2| + 10 log |x-3| + C.

\(\int \frac{\left(x^2+1\right)}{\left(x^2-5 x+6\right)} d x\) = X = 5log |x-2| + 10 log |x-3| + C.

Step-by-Step Guide to Partial Fractions

Example 3 Evaluate \(\int \frac{(3 x-2)}{(x+1)^2(x+3)} d x\)

Solution

Given

\(\int \frac{(3 x-2)}{(x+1)^2(x+3)} d x\)

Let \(\frac{(3 x-2)}{(x+1)^2(x+3)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^2}+\frac{C}{(x+3)}\)

⇒ (3x-2) ≡ A(x+1)(x+3) + B(x+3) + C(x+1)2 …(1)

Putting x = -3 on both sides of (1), we get C = \(-\frac{11}{4}\).

Putting x = -1 on both sides of (1), we get B = \(-\frac{5}{2}\).

Comparing coefficients of x² on both sides of (1), we get

A + C = 0 ⇒ A = -C = \(\frac{11}{4}\).

∴ \(\frac{(3 x-2)}{(x+1)^2(x+3)}=\frac{11}{4(x+1)}-\frac{5}{2(x+1)^2}-\frac{11}{4(x+3)}\)

⇒ \(\int \frac{(3 x-2)}{(x+1)^2(x+3)} d x=\frac{11}{4} \cdot \int \frac{d x}{(x+1)}-\frac{5}{2} \cdot \int \frac{1}{(x+1)^2} d x-\frac{11}{4} \cdot \int \frac{d x}{(x+3)}\)

= \(\frac{11}{4} \cdot \log |x+1|+\frac{5}{2(x+1)}-\frac{11}{4} \cdot \log |x+3|+C\)

= \(\frac{11}{4} \cdot \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C\)

\(\int \frac{(3 x-2)}{(x+1)^2(x+3)} d x\) = \(\frac{11}{4} \cdot \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C\)

Example 4 Evaluate \(\int \frac{d x}{\left(x^3+x^2+x+1\right)}\).

Solution

Given

\(\int \frac{d x}{\left(x^3+x^2+x+1\right)}\)

We have \(\frac{1}{\left(x^3+x^2+x+1\right)}=\frac{1}{x^2(x+1)+(x+1)}=\frac{1}{(x+1)\left(x^2+1\right)} .\)

Let \(\frac{1}{(x+1)\left(x^2+1\right)}=\frac{A}{(x+1)}+\frac{B x+C}{\left(x^2+1\right)}\)

Integration By Partial Fractions Class 12

⇒ 1 ≡ A(x²+1) + (Bx+C)(x+1) …(1)

Putting x = -1 on both sides of (1), we get A = \(\frac{1}{2}\).

Comparing coefficients of x² on both sides of (1), we get

A + B = 0 ⇒ B = -A = \(-\frac{1}{2}\).

Comparing coefficients of x on both sides of (1), we get

B + C = 0 ⇒ C = -B = \(\frac{1}{2}\).

∴ \(\frac{1}{(x+1)\left(x^2+1\right)}=\frac{1}{2(x+1)}+\frac{\frac{-1}{2} x+\frac{1}{2}}{x^2+1}\)

∴ \(\int \frac{d x}{\left(x^3+x^2+x+1\right)}=\int \frac{d x}{(x+1)\left(x^2+1\right)}\)

= \(\frac{1}{2} \cdot \int \frac{d x}{(x+1)}-\frac{1}{2} \int \frac{x}{\left(x^2+1\right)} d x+\frac{1}{2} \int \frac{d x}{\left(x^2+1\right)}\)

= \(\frac{1}{2} \cdot \int \frac{d x}{(x+1)}-\frac{1}{4} \cdot \int \frac{2 x}{\left(x^2+1\right)} d x+\frac{1}{2} \int \frac{d x}{\left(x^2+1\right)}\)

= \(\frac{1}{2} \log |x+1|-\frac{1}{4} \log \left|x^2+1\right|+\frac{1}{2} \tan ^{-1} x+C .\)

\(\int \frac{d x}{\left(x^3+x^2+x+1\right)}\) = \(\frac{1}{2} \log |x+1|-\frac{1}{4} \log \left|x^2+1\right|+\frac{1}{2} \tan ^{-1} x+C .\)

Common Types of Rational Functions for Partial Fractions

Example 5 Evaluate \(\int \frac{x^4}{(x-1)\left(x^2+1\right)} d x \text {. }\)

Solution

Given

\(\int \frac{x^4}{(x-1)\left(x^2+1\right)} d x \text {. }\) \(\frac{x^4}{(x-1)\left(x^2+1\right)}=\frac{x^4}{\left(x^3-x^2+x-1\right)}=(x+1)+\frac{1}{\left(x^3-x^2+x-1\right)}\)

⇒ \(\frac{x^4}{(x-1)\left(x^2+1\right)}=(x+1)+\frac{1}{(x-1)\left(x^2+1\right)}\) …(1)

Let \(\frac{1}{(x-1)\left(x^2+1\right)}=\frac{A}{(x-1)}+\frac{B x+C}{\left(x^2+1\right)}\). Then,

1 ≡ A(x²+1) + (Bx+C)(x-1) …(2)

Putting x = 1 in (2), we get A = \(\frac{1}{2}\).

Comparing coefficients of x² on both sides of (2), we get

A + B = 0 ⇒ B = -A = \(-\frac{1}{2}\).

Comparing the constant terms on both sides of (2), we get

A – C = 1 ⇒ C = (A-1) (\(\frac{1}{2}\) – 1) = \(-\frac{1}{2}\).

∴ \(\frac{1}{(x-1)\left(x^2+1\right)}=\frac{1}{2(x-1)}+\frac{-\frac{1}{2} x-\frac{1}{2}}{\left(x^2+1\right)}\)

∴ \(\frac{x^4}{(x-1)\left(x^2+1\right)}=(x+1)+\frac{1}{2(x-1)}-\frac{1}{2} \cdot \frac{(x+1)}{\left(x^2+1\right)}\)

Integration By Partial

⇒ \(\int \frac{x^4}{(x-1)\left(x^2+1\right)} d x=\int(x+1) d x+\frac{1}{2} \int \frac{d x}{(x-1)}-\frac{1}{4} \cdot \int \frac{2 x}{\left(x^2+1\right)} d x-\frac{1}{2} \int \frac{d x}{\left(x^2+1\right)}\)

= \(\frac{x^2}{2}+x+\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left(x^2+1\right)-\frac{1}{2} \tan ^{-1} x+C\).

\(\int \frac{x^4}{(x-1)\left(x^2+1\right)} d x \text {. }\) = \(\frac{x^2}{2}+x+\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left(x^2+1\right)-\frac{1}{2} \tan ^{-1} x+C\).

Example 6 Evaluate \(\int \frac{(3 x+5)}{\left(x^3-x^2-x+1\right)} d x\)

Solution

Given

\(\int \frac{(3 x+5)}{\left(x^3-x^2-x+1\right)} d x\)

(x³-x²-x+1) = x²(x-1)-(x-1) = (x-1)(x²-1) = (x-1)²(x+1).

Let \(\frac{3 x+5}{\left(x^3-x^2-x+1\right)}=\frac{3 x+5}{(x-1)^2(x+1)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)}\)

⇒ (3x+5) ≡ A(x-1)(x+1) + B(x+1) + C(x-1)² …(1)

Putting x = 1 on both sides of (1), we get B = 4.

Putting x = -1 on both sides of (1), we get C = \(\frac{1}{2}\).

Comparing coefficients of x² on both sides of (1), we get

A + C = 0 ⇒ A = -C = \(-\frac{1}{2}\).

∴ \(\frac{(3 x+5)}{\left(x^3-x^2-x+1\right)}=\frac{-1}{2(x-1)}+\frac{4}{(x-1)^2}+\frac{1}{2(x+1)}\)

⇒ \(\int \frac{(3 x+5)}{\left(x^3-x^2-x+1\right)} d x=-\frac{1}{2} \int \frac{d x}{(x-1)}+4 \int \frac{d x}{(x-1)^2}+\frac{1}{2} \int \frac{d x}{(x+1)}\)

= \(-\frac{1}{2} \log |x-1|-\frac{4}{(x-1)}+\frac{1}{2} \log |x+1|+C .\)

\(\int \frac{(3 x+5)}{\left(x^3-x^2-x+1\right)} d x\) = \(-\frac{1}{2} \log |x-1|-\frac{4}{(x-1)}+\frac{1}{2} \log |x+1|+C .\)

Integration By Partial

Example 7 Evaluate \(\int \frac{\left(x^3-1\right)}{\left(x^3+x\right)} d x .\)

Solution

Given

\(\int \frac{\left(x^3-1\right)}{\left(x^3+x\right)} d x .\)

We have

\(\frac{\left(x^3-1\right)}{\left(x^3+x\right)}=1-\frac{(x+1)}{\left(x^3+x\right)}\) [on dividing]

= \(1-\frac{(x+1)}{x\left(x^2+1\right)}\) …(1)

Let \(\frac{(x+1)}{x\left(x^2+1\right)}=\frac{A}{x}+\frac{B x+C}{\left(x^2+1\right)} .\)

Then, (x+1) ≡ A(x²+1) + (Bx+C)x …(2)

Putting x = 0 in (2), we get A = 1.

Comparing coefficients of x in (2), we get C = 1.

Comparing coefficients of x² in (2), we get

A + B = 0 ⇒ B = -A = -1.

∴ A = 1, B = -1 and C = 1.

Thus, \(\frac{(x+1)}{x\left(x^2+1\right)}=\frac{1}{x}+\frac{(1-x)}{\left(x^2+1\right)}\)

⇒ \(\int \frac{\left(x^3-1\right)}{\left(x^3+x\right)} d x=\int d x-\int \frac{(x+1)}{x\left(x^2+1\right)} d x\)

= \(x-\left\{\int \frac{d x}{x}+\int \frac{(1-x)}{\left(x^2+1\right)} d x\right\}\)

= \(x-\int \frac{d x}{x}-\int \frac{d x}{\left(x^2+1\right)}+\frac{1}{2} \int \frac{2 x}{\left(x^2+1\right)} d x\)

= \(x-\log |x|-\tan ^{-1} x+\frac{1}{2} \log \left(x^2+1\right)+C\).

\(\int \frac{\left(x^3-1\right)}{\left(x^3+x\right)} d x .\) = \(x-\log |x|-\tan ^{-1} x+\frac{1}{2} \log \left(x^2+1\right)+C\).

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Example 8 Evaluate \(\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x\)

Solution

Given

\(\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x\)

Putting sin x = t and cos x dx = dt, we get

Let \(\frac{1}{(1-t)(2-t)}=\frac{A}{(1-t)}+\frac{B}{(2-t)}\)

⇒ 1 ≡ A(2-t) + B(1-t) …(1)

Putting t = 1 in (1), we get A = 1.

Putting t = 2 in (1), we get B = -1.

∴ \(\frac{1}{(1-t)(2-t)}=\frac{1}{(1-t)}-\frac{1}{(2-t)}\)

⇒ \(\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x=\int \frac{d t}{(1-t)(2-t)}\)

= \(\int\left\{\frac{1}{(1-t)}-\frac{1}{(2-t)}\right\} d t\)

= \(\int \frac{d t}{(1-t)}-\int \frac{d t}{(2-t)}\)

= -log |1-t| + log |2-t| + C

= \(\log \left|\frac{2-t}{1-t}\right|+C=\log \left|\frac{2-\sin x}{1-\sin x}\right|+C .\)

\(\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x\) = \(\log \left|\frac{2-t}{1-t}\right|+C=\log \left|\frac{2-\sin x}{1-\sin x}\right|+C .\)

Partial Fractions with Repeated Factors

Example 9 Evaluate \(\int \frac{d x}{x\left\{6(\log x)^2+7 \log x+2\right\}} \text {. }\)

Solution

Given

\(\int \frac{d x}{x\left\{6(\log x)^2+7 \log x+2\right\}} \text {. }\)

Putting log x = t and \frac{1}{x} dx = dt, we get

I = \(\int \frac{d x}{x\left\{6(\log x)^2+7 \log x+2\right\}}=\int \frac{d t}{\left(6 t^2+7 t+2\right)}=\int \frac{d t}{(2 t+1)(3 t+2)} .\)

Let \(\frac{1}{(2 t+1)(3 t+2)}=\frac{A}{(2 t+1)}+\frac{B}{(3 t+2)} \text {. }\)

Then, 1 ≡ A(3t+2) + B(2t+1) …(1)

Putting t = \(-\frac{1}{2}\) in (1), we get A = 2.

Putting t = \(-\frac{2}{3}\) in (1), we get B = -3.

∴ \(\frac{1}{(2 t+1)(3 t+2)}=\frac{2}{(2 t+1)}-\frac{3}{(3 t+2)}\)

⇒ I = \(\int \frac{d t}{(2 t+1)(3 t+2)}\)

= \(\int \frac{2 d t}{(2 t+1)}-\int \frac{3 d t}{(3 t+2)}\)

= log |2t+1| – log |3t+2| + C

= \(\log \left|\frac{2 t+1}{3 t+2}\right|+C\)

= \(\log \left|\frac{2 \log x+1}{3 \log x+2}\right|+C\)

\(\int \frac{d x}{x\left\{6(\log x)^2+7 \log x+2\right\}} \text {. }\) = \(\log \left|\frac{2 \log x+1}{3 \log x+2}\right|+C\)

Example 10 Evaluate \(\int \frac{x^2}{\left(1+x^3\right)\left(2+x^3\right)} d x\)

Solution

Given

\(\int \frac{x^2}{\left(1+x^3\right)\left(2+x^3\right)} d x\)

Putting x³ = t and x²dx = \(\frac{1}{3} dt\), we get

I = \(\int \frac{x^2}{\left(1+x^3\right)\left(2+x^3\right)} d x=\frac{1}{3} \cdot \int \frac{d t}{(1+t)(2+t)} .\)

Let \(\frac{1}{(1+t)(2+t)}=\frac{A}{(1+t)}+\frac{B}{(2+t)}\). Then,

1 ≡ A(2+t) + B(1+t) …(1)

Putting t = -1 in (1), we get A = 1.

Putting t = -2 in (1), we get B = -1.

∴ \(\frac{1}{(1+t)(2+t)}=\frac{1}{(1+t)}-\frac{1}{(2+t)}\)

⇒ I = \(\int \frac{d t}{(1+t)(2+t)}=\int \frac{d t}{(1+t)}-\int \frac{d t}{(2+t)}\)

= log |1+t| – log |2+t| + C

= \(\log \left|\frac{1+t}{2+t}\right|+C\)

= \(\log \left|\frac{1+x^3}{2+x^3}\right|+C\)

\(\int \frac{x^2}{\left(1+x^3\right)\left(2+x^3\right)} d x\) = \(\log \left|\frac{1+x^3}{2+x^3}\right|+C\)

Example 11 Evaluate \(\int \frac{d x}{\left(e^x-1\right)}\)

Solution

Given

\(\int \frac{d x}{\left(e^x-1\right)}\)

Put e^x = t and e^xdx = dt, i.e., dx = \(\frac{1}{t}dt\), we get

I = \(\int \frac{d x}{\left(e^x-1\right)}=\int \frac{d t}{t(t-1)}\)

Let \(\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{(t-1)} .\)

Then, 1 ≡ A(t-1) + Bt …(1)

Putting t = 0 in (1), we get A = -1.

Putting t = 1 in (1), we get B = 1.

∴ \(\frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{(t-1)}\)

Hence, I = \(\int \frac{d x}{\left(e^x-1\right)}\)

= \(\int \frac{d t}{t(t-1)}=\int \frac{-1}{t} d t+\int \frac{1}{(t-1)} d t\)

= -log |t| + log |t-1| + C

= \(\log \left|\frac{t-1}{t}\right|+C\)

= \(\log \left|\frac{e^x-1}{e^x}\right|+C\)

\(\int \frac{d x}{\left(e^x-1\right)}\) = \(\log \left|\frac{e^x-1}{e^x}\right|+C\)

Example 12 Evaluate \(\int \frac{d x}{x\left(x^n+1\right)}\)

Solution

Given

\(\int \frac{d x}{x\left(x^n+1\right)}\)

Putting xⁿ = t, we get nx^n-1dx = dt.

∴ \(\frac{n x^n}{x}\) d x=d t ⇒ \(\frac{1}{x} d x=\frac{1}{n t} d t\) (note)

∴ \(\int \frac{d x}{x\left(x^n+1\right)}=\int \frac{d t}{n t(t+1)}=\frac{1}{n} \cdot \int \frac{d t}{t(t+1)}\) …(1)

Let \(\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{(t+1)} .\)

Then, 1 ≡ A(t+1) + Bt …(2)

Putting t = 0 in (1), we get A = 1.

Putting t = -1 in (1), we get B = -1.

∴ \(\frac{1}{t(t+1)}=\left\{\frac{1}{t}-\frac{1}{(t+1)}\right\}\)

⇒ \(\int \frac{d x}{x\left(x^n+1\right)}=\frac{1}{n} \int \frac{d t}{t(t+1)}\)

= \(\frac{1}{n}\left[\int \frac{1}{t} d t-\int \frac{1}{(t+1)} d t\right]\)

= \(\frac{1}{n} \cdot\{\log |t|-\log |t+1|\}+C\)

= \(\frac{1}{n} \cdot \log \left|\frac{t}{t+1}\right|+C\)

= \(\frac{1}{n} \log \left|\frac{x^n}{x^n+1}\right|+C .\)

\(\int \frac{d x}{x\left(x^n+1\right)}\) = \(\frac{1}{n} \log \left|\frac{x^n}{x^n+1}\right|+C .\)

Applications of Partial Fraction Decomposition in Calculus .

Example 13 Evaluate \(\int \frac{d x}{x\left(x^4+1\right)}\)

Solution

Given

\(\int \frac{d x}{x\left(x^4+1\right)}\)

We have

I = \(\int \frac{d x}{x\left(x^4+1\right)}=\int \frac{x^3}{x^4\left(x^4+1\right)} d x\) [multiplying num. and denom. by x³].

Putting x⁴ = t and 4x³dx = dt, we get

I = \(\frac{1}{4} \cdot \int \frac{d t}{t(t+1)}\)

= \(\frac{1}{4} \cdot \int\left\{\frac{1}{t}-\frac{1}{(t+1)}\right\} d t\) [by partial fraction]

= \(\frac{1}{4} \int \frac{1}{t} d t-\frac{1}{4} \int \frac{1}{(t+1)} d t\)

= \(\frac{1}{4} \log |t|-\frac{1}{4} \log |t+1|+C\)

= \(\frac{1}{4} \log \left|x^4\right|-\frac{1}{4} \log \left|x^4+1\right|+C\)

= \(\left(\frac{1}{4} \times 4\right) \log |x|-\frac{1}{4} \log \left|x^4+1\right|+C\)

= \(\log |x|-\frac{1}{4} \log \left|x^4+1\right|+C\)

\(\int \frac{d x}{x\left(x^4+1\right)}\) = \(\log |x|-\frac{1}{4} \log \left|x^4+1\right|+C\)

Example 14 Evaluate \(\int \frac{x^2}{\left(x^2+2\right)\left(x^2+3\right)} d x\)

Solution

Given

\(\int \frac{x^2}{\left(x^2+2\right)\left(x^2+3\right)} d x\)

Let \(\frac{x^2}{\left(x^2+2\right)\left(x^2+3\right)}=\frac{y}{(y+2)(y+3)}\), where x2 = y.

Let \(\frac{y}{(y+2)(y+3)}=\frac{A}{(y+2)}+\frac{B}{(y+3)}\)

⇒ y ≡ A(y+3) + B(y+2) ..(1)

Putting y = -2 on both sides of (1), we get A = -2.

Putting y = -3 on both sides of (1), we get B = 3.

∴ \(\frac{y}{(y+2)(y+3)}=\frac{-2}{(y+2)}+\frac{3}{(y+3)}\)

⇒ \(\frac{x^2}{\left(x^2+2\right)\left(x^2+3\right)}=\frac{-2}{\left(x^2+2\right)}+\frac{3}{\left(x^2+3\right)}\)

⇒ \(\int \frac{x^2}{\left(x^2+2\right)\left(x^2+3\right)} d x=-2 \cdot \int \frac{d x}{\left(x^2+2\right)}+3 \cdot \int \frac{d x}{\left(x^2+3\right)}\)

= \(\frac{-2}{\sqrt{2}} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{3}{\sqrt{3}} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)+C\)

= \(-\sqrt{2} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)+\sqrt{3} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)+C\)

\(\int \frac{x^2}{\left(x^2+2\right)\left(x^2+3\right)} d x\) = \(-\sqrt{2} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)+\sqrt{3} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)+C\)

Example 15 Evaluate \(\int \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)} d x\)

Solution

Given

\(\int \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)} d x\)

Let \(\frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)}=1+\frac{A}{(x-4)}+\frac{B}{(x-5)}+\frac{C}{(x-6)}\). Then,

(x-1)(x-2)(x-3) ≡ (x-4)(x-5)(x-6) + A(x-5)(x-6) + B(x-4)(x-6) + C(x-4)(x-5) …(1)

Putting x = 4 on both sides of (1) we get A = 3.

Putting x = 5 on both sides of (1), we get B = -24.

Putting x = 6 on both sides of (1), we get C = 30.

∴ I = \(\int \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)} d x\)

= \(\int\left\{1+\frac{3}{(x-4)}-\frac{24}{(x-5)}+\frac{30}{(x-6)}\right\} d x\)

= \(\int d x+3 \int \frac{d x}{(x-4)}-24 \int \frac{d x}{(x-5)}+30 \int \frac{d x}{(x-6)}\)

= x + 3 log |x-4| – 24 log |x-5| + 30 log |x-6| + C.

\(\int \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)} d x\) = x + 3 log |x-4| – 24 log |x-5| + 30 log |x-6| + C.

Example 16 Evaluate \(\int \frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)} d x\)

Solution

Given

\(\int \frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)} d x\)

We have

\(\frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)}=\frac{(t+1)(t+2)}{(t+3)(t+4)}\), where x² = t

= \(\frac{\left(t^2+3 t+2\right)}{\left(t^2+7 t+12\right)}=1-\frac{(4 t+10)}{(t+3)(t+4)}\).

Let \(\frac{(4 t+10)}{(t+3)(t+4)}=\frac{A}{(t+3)}+\frac{B}{(t+4)}\)

⇒ (4t+10) ≡ A(t+4) + B(t+3) …(1)

Putting t =- -3 in (1), we get A = -2.

Putting t = -4 in (1), we get B = 6.

∴ \(\frac{(4 t+10)}{(t+3)(t+4)}=\frac{-2}{(t+3)}+\frac{6}{(t+4)}\) …(2)

Thus, \(\frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)}=\frac{(t+1)(t+2)}{(t+3)(t+4)}\), where x2 = t-1

= \(\frac{\left(t^2+3 t+2\right)}{\left(t^2+7 t+12\right)}=1-\frac{(4 t+10)}{(t+3)(t+4)}\)

= \(1-\left\{\frac{-2}{(t+3)}+\frac{6}{(t+4)}\right\}\) [from (2)]

= \(\left\{1+\frac{2}{(t+3)}-\frac{6}{(t+4)}\right\}=\left\{1+\frac{2}{\left(x^2+3\right)}-\frac{6}{\left(x^2+4\right)}\right\}\)

∴ \(\int \frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)} d x=\int\left\{1+\frac{2}{\left(x^2+3\right)}-\frac{6}{\left(x^2+4\right)}\right\} d x\)

= \(\int d x+2 \int \frac{d x}{\left(x^2+3\right)}-6 \int \frac{d x}{\left(x^2+4\right)}\)

= \(x+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)-\frac{6}{2} \tan ^{-1}\left(\frac{x}{2}\right)+C\)

= \(x+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)-3 \tan ^{-1}\left(\frac{x}{2}\right)+C\).

\(\int \frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)} d x\) = \(x+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)-3 \tan ^{-1}\left(\frac{x}{2}\right)+C\).

Partial Fraction Integration Techniques for Beginners

Example 17 Evaluate \(\int \frac{(3 \sin \theta-2) \cos \theta}{\left(5-\cos ^2 \theta-4 \sin \theta\right)} d \theta\)

Solution

Given:

\(\int \frac{(3 \sin \theta-2) \cos \theta}{\left(5-\cos ^2 \theta-4 \sin \theta\right)} d \theta\)

We have

I = \(\int \frac{(3 \sin \theta-2) \cos \theta}{\left\{5-\left(1-\sin ^2 \theta\right)-4 \sin \theta\right\}} d \theta\)

= \(\int \frac{(3 \sin \theta-2) \cos \theta}{\left(4+\sin ^2 \theta-4 \sin \theta\right)} d \theta\)

= \(\int \frac{(3 \sin \theta-2) \cos \theta}{(\sin \theta-2)^2} d \theta=\int \frac{(3 t-2)}{(t-2)^2} d t\), where sin θ = t.

Let \(\frac{(3 t-2)}{(t-2)^2}=\frac{A}{(t-2)}+\frac{B}{(t-2)^2}\). Then,

(3t-2) ≡ A(t-2) + B …(1)

Putting t = 2 in (1), we get B = 4.

Comparing coefficients of t on both sides of (1), we get A = 3.

Thus, A = 3 and B = 4.

∴ \(\frac{(3 t-2)}{(t-2)^2}=\frac{3}{(t-2)}+\frac{4}{(t-2)^2}\)

⇒ I = \(\int \frac{(3 t-2)}{(t-2)^2} d t=\int \frac{3}{(t-2)} d t+\int \frac{4}{(t-2)^2} d t\)

= \(3 \log |t-2|-\frac{4}{(t-2)}+C\)

= \(3 \log |\sin \theta-2|-\frac{4}{(\sin \theta-2)}+C\)

= \(3 \log (2-\sin \theta)+\frac{4}{(2-\sin \theta)}+C\) [∵ (2-sin θ) > 0].

\(\int \frac{(3 \sin \theta-2) \cos \theta}{\left(5-\cos ^2 \theta-4 \sin \theta\right)} d \theta\) = \(3 \log (2-\sin \theta)+\frac{4}{(2-\sin \theta)}+C\) [∵ (2-sin θ) > 0].

Example 18 Evaluate \(\int \frac{\left(\tan \theta+\tan ^3 \theta\right)}{\left(1+\tan ^3 \theta\right)} d \theta\)

Solution

Given

\(\int \frac{\left(\tan \theta+\tan ^3 \theta\right)}{\left(1+\tan ^3 \theta\right)} d \theta\)

We have

\(\frac{\left(\tan \theta+\tan ^3 \theta\right)}{\left(1+\tan ^3 \theta\right)}=\frac{\tan \theta\left(1+\tan ^2 \theta\right)}{\left(1+\tan ^3 \theta\right)}=\frac{\tan \theta \sec ^2 \theta}{\left(1+\tan ^3 \theta\right)} .\)

∴ I = \(\int \frac{\left(\tan \theta+\tan ^3 \theta\right)}{\left(1+\tan ^3 \theta\right)} d \theta\)

= \(\int \frac{\tan \theta \sec ^2 \theta}{\left(1+\tan ^3 \theta\right)} d \theta\)

= \(\int \frac{t}{\left(1+t^3\right)} d t=\int \frac{t}{(1+t)\left(1-t+t^2\right)} d t\), where tan θ = t.

Let \(\frac{t}{(1+t)\left(1-t+t^2\right)}=\frac{A}{(1+t)}+\frac{(B t+C)}{\left(1-t+t^2\right)}\). Then,

t ≡ A(1-t+t²) + (Bt+C)(1+t) …(1)

Putting t = -1 on both sides of (1), we get A = \(-\frac{1}{3}\).

Comparing coefficients of t2 on both sides of (1), we get

A + B = 0 ⇒ B = -A = \(\frac{1}{3}\).

Comparing constant terms on both sides of (1), we get

A + C = 0 ⇒ C = -A = \(\frac{1}{3}\).

∴ \(\frac{t}{(1+t)\left(1-t+t^2\right)}=\frac{-1}{3(1+t)}+\frac{\left(\frac{1}{3} t+\frac{1}{3}\right)}{\left(1-t+t^2\right)}\)

Now, I = \(\int \frac{t}{(1+t)\left(1-t+t^2\right)} d t\)

= \(-\frac{1}{3} \int \frac{d t}{(1+t)}+\frac{1}{6} \int \frac{2 t}{\left(t^2-t+1\right)} d t+\frac{1}{3} \int \frac{d t}{\left(t^2-t+1\right)}\)

= \(-\frac{1}{3} \int \frac{d t}{(1+t)}+\frac{1}{6} \int \frac{(2 t-1)+1}{\left(t^2-t+1\right)} d t+\frac{1}{3} \int \frac{d t}{\left(t^2-t+1\right)}\)

= \(-\frac{1}{3} \int \frac{d t}{(1+t)}+\frac{1}{6} \int \frac{(2 t-1)}{\left(t^2-t+1\right)} d t+\frac{1}{2} \int \frac{d t}{\left(t^2-t+1\right)}\)

= \(-\frac{1}{3} \log |1+t|+\frac{1}{6} \log \left|t^2-t+1\right|+\frac{1}{2} \int \frac{d t}{\left(t^2-t+\frac{1}{4}\right)+\frac{3}{4}}\)

= \(-\frac{1}{3} \log |1+t|+\frac{1}{6} \log \left|t^2-t+1\right|+\frac{1}{2} \int \frac{d t}{(t-1 / 2)^2+(\sqrt{3} / 2)^2}\)

= \(-\frac{1}{3} \log |1+t|+\frac{1}{6} \log \left|t^2-t+1\right|+\frac{1}{2} \cdot \frac{2}{\sqrt{3}} \tan ^{-1} \frac{\left(t-\frac{1}{2}\right)}{(\sqrt{3} / 2)}+C\)

Comparative Analysis of Different Partial Fraction Forms

= \(-\frac{1}{3} \log |1+t|+\frac{1}{6} \log \left|t^2-t+1\right|+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t-1}{\sqrt{3}}\right)+C\)

= \(-\frac{1}{3} \log |1+\tan \theta|+\frac{1}{6} \log \left|\tan ^2 \theta-\tan \theta+1\right|+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan \theta-1}{\sqrt{3}}\right)+C .\)

\(\int \frac{\left(\tan \theta+\tan ^3 \theta\right)}{\left(1+\tan ^3 \theta\right)} d \theta\) = \(-\frac{1}{3} \log |1+\tan \theta|+\frac{1}{6} \log \left|\tan ^2 \theta-\tan \theta+1\right|+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan \theta-1}{\sqrt{3}}\right)+C .\)

Example 19 Evaluate \(\int \frac{d x}{(\sin x-\sin 2 x)}\).

Solution

Given

\(\int \frac{d x}{(\sin x-\sin 2 x)}\) \(\int \frac{d x}{(\sin x-\sin 2 x)}=\int \frac{d x}{(\sin x-2 \sin x \cos x)}\)

= \(\int \frac{d x}{\sin x(1-2 \cos x)}=\int \frac{\sin x}{\sin ^2 x(1-2 \cos x)} d x\)

= \(\int \frac{\sin x}{\left(1-\cos ^2 x\right)(1-2 \cos x)} d x\)

= \(-\int \frac{d t}{\left(1-t^2\right)(1-2 t)}\), where cos x = t

= \(\int \frac{d t}{(t-1)(t+1)(1-2 t)}\) …(1)

Let \(\frac{1}{(t-1)(t+1)(1-2 t)}=\frac{A}{(t-1)}+\frac{B}{(t+1)}+\frac{C}{(1-2 t)} .\)

Then, 1 ≡ A(t+1)(1-2t) + B(t-1)(1-2t) + C(t-1)(t+1)  …(2)

Putting t = 1 in (2), we get A = \(-\frac{1}{2}\).

Putting t = -1 in (2), we get B = \(-\frac{1}{6}\).

Putting t = \(\frac{1}{2}\) in (2), we get C = \(-\frac{4}{3}\).

∴ I = \(-\frac{1}{2} \int \frac{d t}{(t-1)}-\frac{1}{6} \cdot \int \frac{d t}{(t+1)}-\frac{4}{3} \cdot \int \frac{d t}{(1-2 t)}\)

= \(-\frac{1}{2} \log |t-1|-\frac{1}{6} \log |t+1|+\frac{2}{3} \cdot \int \frac{-2 d t}{(1-2 t)}\)

= \(-\frac{1}{2} \log |t-1|-\frac{1}{6} \log |t+1|+\frac{2}{3} \log |1-2 t|+C\)

= \(-\frac{1}{2} \log |\cos x-1|-\frac{1}{6} \log |\cos x+1|+\frac{2}{3} \log |1-2 \cos x|+C\) .

\(\int \frac{d x}{(\sin x-\sin 2 x)}\) = \(-\frac{1}{2} \log |\cos x-1|-\frac{1}{6} \log |\cos x+1|+\frac{2}{3} \log |1-2 \cos x|+C\) .

Example 20 Evaluate \(\int \frac{(1-\cos x)}{\cos x(1+\cos x)} d x\).

Solution

Given

\(\int \frac{(1-\cos x)}{\cos x(1+\cos x)} d x\)

Let \(\frac{(1-\cos x)}{\cos x(1+\cos x)}=\frac{(1-t)}{t(1+t)}=\frac{A}{t}+\frac{B}{(1+t)}\), where t = cos x.

Then, (1-t) ≡ A(1+t) + Bt.

Integration By Partial Fractions Class 12

Putting t = 0 in this identity, we get A = 1.

Putting t = -1 in this identity, we get B = -2.

∴ \(\frac{(1-t)}{t(1+t)}=\frac{1}{t}-\frac{2}{1+t}\)

or \(\frac{(1-\cos x)}{\cos x(1+\cos x)}=\frac{1}{\cos x}-\frac{2}{(1+\cos x)}\)

∴ \(\int \sec x d x-\int \sec ^2 \frac{x}{2} d x\)

= \(\log |\sec x+\tan x|-2 \tan \frac{x}{2}+C\).

\(\int \frac{(1-\cos x)}{\cos x(1+\cos x)} d x\) = \(\log |\sec x+\tan x|-2 \tan \frac{x}{2}+C\).

Example 21 Evaluate \(\int \frac{\left(x^2+1\right)}{(x+1)^2} d x\).

Solution

Given

\(\int \frac{\left(x^2+1\right)}{(x+1)^2} d x\)

On dividing (x²+1) by (x²+2x+1), we get

\(\frac{\left(x^2+1\right)}{(x+1)^2}=\left\{1-\frac{2 x}{(x+1)^2}\right\} .\)

Let \(\frac{2 x}{(x+1)^2}=\frac{A}{(x+1)}+\frac{B}{(x+1)^2}\)

⇒ 2x ≡ A(x+1) + B …(1)

On equating the coefficients of x, we get A = 2.

On equating constant terms, we get A + B = 0 ⇒ B = -A = -2.

∴ \(\frac{2 x}{(x+1)^2}=\frac{2}{(x+1)}-\frac{2}{(x+1)^2}\)

∴ I = \(\int\left\{1-\frac{2 x}{(x+1)^2}\right\}\)

= \(\int\left\{1-\frac{2}{(x+1)}+\frac{2}{(x+1)^2}\right\} d x\)

= \(x-2 \log |x+1|-\frac{2}{(x+1)}+C .\)

\(\int \frac{\left(x^2+1\right)}{(x+1)^2} d x\) = \(x-2 \log |x+1|-\frac{2}{(x+1)}+C .\)

Integration In Mathematics

Integration Types – Methods Of Integration

Integration by Substitution

If we have to evaluate an integral of the type \(\int f\{\phi(x)\} \cdot \phi^{\prime}(x) d x\) then we put Φ(x) = t and Φ'(x)dx = dt. With this substitution, the integrand becomes easily integrable.

Case 1 When the integrand is of the form f(ax+b), we put (ax+b) = t and \(d x=\frac{1}{a} d t .\)

Case 2 When the integrand is of the form x^n-1.f(xⁿ), we put xⁿ = t and nx^n-1dx = dt.

Case 3 When the integrand is of the form {f(x)}ⁿ.f'(x), we put f(x) = t and f'(x) dx = dt .

Case 4 When the integrand is of the form \(\frac{f^{\prime}(x)}{f(x)}\), we put f(x) = t and f'(x)dx = dt.

Theorem 1 \(\int(a x+b)^n d x=\frac{(a x+b)^{n+1}}{a(n+1)}+C\), where n ≠ -1.

Proof

Putting ax + b = t, we get a dx = dt or dx = \(\frac{1}{a}\)dt.

∴ \(\int(a x+b)^n d x=\frac{1}{a} \int t^n d t\)=\(\frac{1}{a} \cdot \frac{t^{n+1}}{(n+1)}+C\)=\(\frac{(a x+b)^{n+1}}{a(n+1)}+C .\)

Theorem 2

\(\int \cos (a x+b) d x=\frac{1}{a} \sin (a x+b)+C\)

Read and Learn More  Class 12 Math Solutions

Proof

Put (ax+b) = t so that dx = \(\frac{1}{a}\)dt.

∴ \(\int \cos (a x+b) d x=\frac{1}{a} \int \cos t d t\)

= \(\frac{1}{a} \sin t+C\)

= \(\frac{1}{a} \sin (a x+b)+C .\)

Integration Types

Solved Examples

Example 1 Evaluate:

(1) \(\int(3 x+5)^7 d x\)

(2) \(\int(4-9 x)^5 d x\)

(3) \(\int \frac{1}{(2-3 x)^4} d x\)

(4) \(\int \sqrt{a x+b} d x\)

Solution

(1) Put (3x+5) = t so that 3dx = dt or dx = \(\frac{1}{3}\)dt.

∴ \(\int(3 x+5)^7 d x=\frac{1}{3} \int t^7 d t=\frac{1}{3} \cdot \frac{t^8}{8}+C=\frac{(3 x+5)^8}{24}+C .\)

(2) Put (4-9x) = t so that -9dx = dt or dx = \(-\frac{1}{9}\)dt.

∴ \(\int(4-9 x)^5 d x=-\frac{1}{9} \int t^5 d t=-\frac{1}{9} \cdot \frac{t^6}{6}+C=\frac{-(4-9 x)^6}{54}+C\).

(3) Put (2-3x) = t so that -3dx = dt or dx = \(-\frac{1}{9}\)dt.

∴ \(\int \frac{1}{(2-3 x)^4} d x=-\frac{1}{3} \int \frac{1}{t^4} d t=-\frac{1}{3} \cdot \frac{1}{\left(-3 t^3\right)}+C=\frac{1}{9(2-3 x)^3}+C\).

(4) Put (ax+b) = t so that a dx = dt.

∴ \(\int \sqrt{a x+b} d t=\frac{1}{a} \int \sqrt{t} d t=\frac{2}{3 a} t^{3 / 2}+C=\frac{2(a x+b)^{3 / 2}}{3 a}+C .\)

Example 2 Evaluate:

(1) \(\int \cos 2 x d x\)

(2) \(\int e^{(5 x+3)} d x\)

(3) \(\int \sec ^2(3 x+5) d x\)

(4) \(\int \sin ^3 x d x\)

Integration Types

Solution

(1) Put 2x = t so that 2 dx = dt or dx = \(\frac{1}{2}\) dt.

∴ \(\int \cos 2 x d x=\frac{1}{2} \int \cos t d t=\frac{1}{2} \sin t+C=\frac{1}{2} \sin 2 x+C .\)

(2) Put (5x+3) = t so that 5dx = dt or dx = \(\frac{1}{5}\) dt.

∴ \(\int e^{(5 x+3)} d x=\frac{1}{5} \int e^t d t=\frac{1}{5} \cdot e^t+C=\frac{1}{5} e^{(5 x+3)}+C .\)

(3) Put (3x+5) = t so that 3dx = dt or dx = \(\frac{1}{3}\) dt.

∴ \(\int \sec ^2(3 x+5) d x=\frac{1}{3} \int \sec ^2 t d t=\frac{1}{3} \tan t+C\)

= \(\frac{1}{3} \tan (3 x+5)+C\)

(4) We know that sin 3x = 3 sin x – 4 sin³x.

∴ \(\sin ^3 x=\frac{1}{4}(3 \sin x-\sin 3 x)\)

So, \(\int \sin ^3 x d x=\int\left(\frac{3}{4} \sin x-\frac{1}{4} \sin 3 x\right) d x\)

= \(\frac{3}{4} \int \sin x d x-\frac{1}{4} \int \sin 3 x d x\)

= \(\frac{3}{4}(-\cos x)-\frac{1}{4} \cdot \frac{(-\cos 3 x)}{3}+C\)

= \(-\frac{3}{4} \cos x+\frac{\cos 3 x}{12}+C\)

Integration in Mathematics 

Basic Concepts of Integration

Example 3 Evaluate:

(1) \(\int \frac{\log x}{x} d x\)

(2) \(\int \frac{\sec ^2(\log x)}{x} d x\)

(3) \(\int \frac{e^{\tan ^{-1} x}}{\left(1+x^2\right)} d x\)

(4) \(\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x\)

Solution

(1) Put log x = t so that \(\frac{1}{2}\) dx = dt.

∴ \(\int \frac{\log x}{x} d x=\int t d t=\frac{1}{2} t^2+C=\frac{1}{2}(\log x)^2+C\)

Integration Types

(2) Put log x = t so that \(\frac{1}{2}\) dx = dt.

∴ \(\int \frac{\sec ^2(\log x)}{x} d x=\int \sec ^2 t d t=\tan t+C=\tan (\log x)+C .\)

(3) Put tan-1x = t so that \(\frac{1}{\left(1+x^2\right)}\) dx = dt.

∴ \(\int \frac{e^{\tan ^{-1} x}}{\left(1+x^2\right)} d x=\int e^t d t=e^t+C=e^{\tan ^{-1} x}+C\)

(4) Put √x = t so that \(\frac{1}{2} x^{-1 / 2}\) dx = dt

or \(\frac{1}{\sqrt{x}} d x=2 d t\)

∴ \(\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x=2 \int \sin t d t\)

= 2(-cos t) + C = -2 cos T + C = -2 cos √x + C.

Integration Examples

Example 4 Evaluate:

(1) \(\int \cos ^3 x \sin x d x\)

(2) \(\int(\sqrt{\sin x}) \cos x d x\)

(3) \(\int \frac{\sin x}{(3+4 \cos x)^2} d x\)

Solution

(1) Put cos x = t so that sin x dx = -dt.

∴ \(\int \cos ^3 x \sin x d x=-\int t^3 d t=-\frac{t^4}{4}+C=-\frac{1}{4} \cos ^4 x+C \text {. }\)

(2) Put sin x = t so that cos x dx = dt.

∴ \(\int(\sqrt{\sin x}) \cos x d x=\int \sqrt{t} d t=\frac{2}{3} t^{3 / 2}+C=\frac{2}{3}(\sin x)^{3 / 2}+C\).

(3) Put (3 + 4 cos x) = t so that -4sin x dx = dt.

∴ \(\int \frac{\sin x}{(3+4 \cos x)^2} d x=-\frac{1}{4} \int \frac{1}{t^2} d t=\frac{1}{4 t}+C=\frac{1}{4(3+4 \cos x)}+C \text {. }\)

Integration Examples

Example 5 Evaluate:

(1) \(\int \frac{2 x}{(2 x+1)^2} d x\)

(2) \(\int \frac{(2+3 x)}{(3-2 x)} d x\)

Solution

(1) Put (2x + 1) = t so that 2x = (t – 1) and dx = \(\frac{1}{2}\) dt.

∴ \(\int \frac{2 x}{(2 x+1)^2} d x=\frac{1}{2} \int \frac{(t-1)}{t^2} d t\)

= \(\frac{1}{2} \int \frac{1}{t} d t-\frac{1}{2} \int \frac{1}{t^2} d t=\frac{1}{2} \log |t|+\frac{1}{2 t}+C\)

= \(\frac{1}{2} \log |(2 x+1)|+\frac{1}{2(2 x+1)}+C .\)

(2) Put (3 – 2x) = t so that x = \(\left(\frac{3-t}{2}\right)\) and dx = \(-\frac{1}{2}\) dt.

∴ \(\int \frac{(2+3 x)}{(3-2 x)} d t=-\frac{1}{2} \int \frac{\left[2+\left(\frac{9-3 t}{2}\right)\right]}{t} d t=-\frac{1}{4} \int \frac{(13-3 t)}{t} d t\)

= \(\frac{-13}{4} \int \frac{1}{t} d t+\frac{3}{4} \int d t=-\frac{13}{4} \log |t|+\frac{3}{4} t+C\)

= \(-\frac{13}{4} \log |(3-2 x)|+\frac{3}{4}(3-2 x)+C .\)

Integration in Mathematics 

Integration Examples

Example 6 Evaluate:

(1) \(\int \frac{3 x^2}{\left(1+x^6\right)} d x\)

(2) \(\int \frac{x^3}{\left(x^2+1\right)^3} d x\)

(3) \(\frac{x^8}{\left(1-x^3\right)^{1 / 3}} d x\)

Solution

(1) Put x³ = t so that 3x²dx = dt.

∴ \(\int \frac{3 x^2}{\left(1+x^6\right)} d x=\int \frac{d t}{\left(1+t^2\right)}=\tan ^{-1} t+C=\tan ^{-1} x^3+C .\)

(2) Put (x²+1) = t so that x² = (t-1) and x dx = \(\frac{1}{2}\) dt.

∴ \(\int \frac{x^3}{\left(x^2+1\right)^3} d x=\int \frac{x^2 \cdot x}{\left(x^2+1\right)^3} d x\)

= \(\frac{1}{2} \int \frac{(t-1)}{t^3} d t=\frac{1}{2} \int \frac{1}{t^2} d t-\frac{1}{2} \int \frac{1}{t^3} d t\)

= \(\frac{-1}{2 t}+\frac{1}{4 t^2}+C=\frac{-1}{2\left(x^2+1\right)}+\frac{1}{4\left(x^2+1\right)^2}+C\)

= \(\frac{-\left(1+2 x^2\right)}{4\left(x^2+1\right)^2}+C\)

(3) Put (1-x³) = t so that x³ = (t-1) and x²dx = \(-\frac{1}{3}\) dt.

∴ \(\int \frac{x^8}{\left(1-x^3\right)^{1 / 3}} d x=\int \frac{x^6 \cdot x^2}{\left(1-x^3\right)^{1 / 3}} d x\)

= \(-\frac{1}{3} \int \frac{(1-t)^2}{t^{1 / 3}} d t=-\frac{1}{3} \int \frac{\left(1+t^2-2 t\right)}{t^{1 / 3}} d t\)

= \(-\frac{1}{3} \int t^{-1 / 3} d t-\frac{1}{3} \int t^{5 / 3} d t+\frac{2}{3} \int t^{2 / 3} d t\)

Integration Examples

= \(-\frac{1}{2} t^{2 / 3}-\frac{1}{8} t^{8 / 3}+\frac{2}{5} t^{5 / 3}+\mathrm{C}\)

= \(-\frac{1}{2}\left(1-x^3\right)^{2 / 3}-\frac{1}{8}\left(1-x^3\right)^{8 / 3}+\frac{2}{5}\left(1-x^3\right)^{5 / 3}+C\)

Types of Integration Methods Explained

Example 7 Evaluate \(\int \frac{d x}{x \cdot \sqrt{x^6-1}}\)

Solution

Given

\(\int \frac{d x}{x \cdot \sqrt{x^6-1}}\)

Put x³ = t so that 3x²dx = dt or x²dx = \(\frac{1}{3}\) dt.

∴ \(\int \frac{d x}{x \cdot \sqrt{x^6-1}}=\int \frac{x^2}{x^3 \cdot \sqrt{x^6-1}} d x\)

[multiplying numerator and denominator by x2]

= \(\frac{1}{3} \int \frac{1}{t \sqrt{t^2-1}} d t=\frac{1}{3} \sec ^{-1} t+C=\frac{1}{3} \sec ^{-1} x^3+C\)

\(\int \frac{d x}{x \cdot \sqrt{x^6-1}}\) = \(\frac{1}{3} \int \frac{1}{t \sqrt{t^2-1}} d t=\frac{1}{3} \sec ^{-1} t+C=\frac{1}{3} \sec ^{-1} x^3+C\)

Example 8 Evaluate \(\int \frac{1}{(\sqrt{x}+x)} d x\)

Solution

Given

\(\int \frac{1}{(\sqrt{x}+x)} d x\) \(\int \frac{1}{(\sqrt{x}+x)} d x=\int \frac{1}{\sqrt{x}(1+\sqrt{x})} d x\)

Now, put (1+√x) = t so that \(\frac{1}{\sqrt{x}} d x\) = 2 dt.

∴ \(\int \frac{1}{(\sqrt{x}+x)} d x=\int \frac{1}{\sqrt{x}(1+\sqrt{x})} d x\)

= \(2 \int \frac{1}{t} d t=2 \log |t|+C=2 \log |(1+\sqrt{x})|+C\).

\(\int \frac{1}{(\sqrt{x}+x)} d x\) = \(2 \int \frac{1}{t} d t=2 \log |t|+C=2 \log |(1+\sqrt{x})|+C\).

Integration in Mathematics 

Example 9 Evaluate:

(1) \(\int \frac{(x-1)}{\sqrt{x+4}} d x\)

(2) \(\int x \sqrt{x+2} d x\)

(3) \(\int(4 x+2) \sqrt{x^2+x+1} d x\)

(4) \(\int \frac{(4 x+3)}{\sqrt{2 x^2+3 x+1}} d x\)

Solution

(1) Put (x+4) = t² so that x = (t² – 4) and dx = 2t dt.

∴ \(\int \frac{(x-1)}{\sqrt{x+4}} d x=2 \int \frac{\left(t^2-5\right) t}{t} d t\)

= \(2 \int t^2 d t-10 \int d t=\frac{2 t^3}{3}-10 t+C\)

= \(\frac{2}{3}(x+4)^{3 / 2}-10(x+4)^{1 / 2}+C\)

(2) Put (x+2) = t² so that x = (t²-2) and dx = 2t dt.

Definition Of Integration In Maths

∴ \(\int x \sqrt{x+2} d x=\int\left(t^2-2\right) 2 t^2 d t=2 \int t^4 d t-4 \int t^2 d t\)

= \(\frac{2 t^5}{5}-\frac{4 t^3}{3}+C=\frac{2(x+2)^{5 / 2}}{5}-\frac{4(x+2)^{3 / 2}}{3}+C\).

(3) Put (x²+x+1) = t so that (2x+1)dx = dt.

∴ \(\int(4 x+2)\left(\sqrt{x^2+x+1}\right) d x=2 \int \sqrt{t} d t\)

= \(\frac{4}{3} t^{3 / 2}+C=\frac{4}{3}\left(x^2+x+1\right)^{3 / 2}+C\).

(4) Put (2x²+3x+1) = t so that (4x+3)dx = dt.

∴ \(\int \frac{(4 x+3)}{\sqrt{2 x^2+3 x+1}} d x=\int \frac{d t}{\sqrt{t}}=2 \sqrt{t}+C=2 \sqrt{2 x^2+3 x+1}+C\)

Example 10 Evaluate:

(1) \(\int \frac{(2 x+5)}{\left(x^2+5 x+9\right)} d x\)

(2) \(\int \frac{(6 x-7)}{\left(3 x^2-7 x+5\right)^2} d x\)

(3) \(\int \frac{(\cos x-\sin x)}{(\cos x+\sin x)} d x\)

(4) \(\int \frac{\sec x}{\log (\sec x+\tan x)} d x\)

Solution

(1) Put (x²+5x+9) = t so that (2x+5)dx = dt.

∴ \(\int \frac{(2 x+5)}{\left(x^2+5 x+9\right)} d x=\int \frac{1}{t} d t=\log |t|+C\)

= log|(x2+5x+9)| + C.

(2) Put (3x²-7x+5) = t so that (6x-7)dx = dt.

Definition Of Integration In Maths

∴ \(\int \frac{(6 x-7)}{\left(3 x^2-7 x+5\right)^2} d x=\int \frac{1}{t^2} d t=-\frac{1}{t}+C=\frac{-1}{\left(3 x^2-7 x+5\right)}+C\)

(3) Put (cos x + sin x) = t so that (cos x – sin x)dx = dt.

∴ \(\int \frac{(\cos x-\sin x)}{(\cos x+\sin x)} d x=\int \frac{1}{t} d t\)

= log |t| + C = log |(cos x + sin x)| + C.

(4) Put log(sec x + tan x) = t.

Then, on differentiation, we get

\(\frac{1}{(\sec x+\tan x)} \cdot\left(\sec x \tan x+\sec ^2 x\right) d x=d t\)

or sec x dx = dt.

∴ \(\int \frac{\sec x}{\log (\sec x+\tan x)} d x=\int \frac{1}{t} d t\)

= log |t| + C = log |log(sec x + tan x)| + C.

Example 11 Evaluate \(\int \frac{\sin 2 x}{\left(a^2 \sin ^2 x+b^2 \cos ^2 x\right)} d x\)

Solution

Given

\(\int \frac{\sin 2 x}{\left(a^2 \sin ^2 x+b^2 \cos ^2 x\right)} d x\)

Put (a² sin²x + b²cos²x) = t so that

\(2(a2 – b2)sin x cos x dx = dt ⇔ sin 2x dx = \frac{d t}{\left(a^2-b^2\right)}.\)

∴ I = \(\int \frac{\sin 2 x}{\left(a^2 \sin ^2 x+b^2 \cos ^2 x\right)} d x=\frac{d t}{\left(a^2-b^2\right) t}\)

= \(\frac{1}{\left(a^2-b^2\right)} \log |t|+C\)

Definition Of Integration In Maths

= \(\frac{1}{\left(a^2-b^2\right)} \log \left(a^2 \sin ^2 x+b^2 \cos ^2 x\right)+C\).

\(\int \frac{\sin 2 x}{\left(a^2 \sin ^2 x+b^2 \cos ^2 x\right)} d x\) = \(\frac{1}{\left(a^2-b^2\right)} \log \left(a^2 \sin ^2 x+b^2 \cos ^2 x\right)+C\).

Integration by Substitution Examples

Example 12 Evaluate \(\int \frac{x^2 \tan ^{-1} x^3}{\left(1+x^6\right)} d x\)

Solution

Given

\(\int \frac{x^2 \tan ^{-1} x^3}{\left(1+x^6\right)} d x\)

Put tan^-1x²³ = t so that \(\frac{3 x^2}{\left(1+x^6\right)} d x=d t or \frac{x^2}{\left(1+x^6\right)} d x=\frac{1}{3} d t\)

∴ \(\int \frac{x^2 \tan ^{-1} x^3}{\left(1+x^6\right)} d x=\frac{1}{3} \int t d t=\frac{1}{6} t^2+C=\frac{1}{6}\left(\tan ^{-1} x^3\right)^2+C \text {. }\)

Example 13 Evaluate \(\int \frac{\sqrt{\tan x}}{\sin x \cos x} d x\)

Solution

Given

\(\int \frac{\sqrt{\tan x}}{\sin x \cos x} d x\) \(\int \frac{\sqrt{\tan x}}{\sin x \cos x} d x=\int \frac{\tan x}{(\sqrt{\tan x}) \cdot \sin x \cos x} d x=\int \frac{\sec ^2 x}{\sqrt{\tan x}} d x\)

= \(\int \frac{1}{\sqrt{t}} d t, where tan x = t and sec2x dx = dt\)

= 2√t + C = 2√tan x + C.

\(\int \frac{\sqrt{\tan x}}{\sin x \cos x} d x\) = 2√t + C = 2√tan x + C.

Types of Integration 

Example 14 Evaluate \(\int \frac{d x}{(\sqrt{2 x+3}+\sqrt{2 x-3})}\)

Solution

Given

\(\int \frac{d x}{(\sqrt{2 x+3}+\sqrt{2 x-3})}\)

= \(\int \frac{1}{(\sqrt{2 x+3}+\sqrt{2 x-3})} \times \frac{(\sqrt{2 x+3})-\sqrt{2 x-3})}{(\sqrt{2 x+3})-\sqrt{2 x-3})} d x\)

= \(\int \frac{(\sqrt{2 x+3}-\sqrt{2 x-3})}{[(2 x+3)-(2 x-3)]} d x=\frac{1}{6} \int(2 x+3)^{1 / 2} d x-\frac{1}{6} \int(2 x-3)^{1 / 2} d x\)

= \(\frac{1}{18}(2 x+3)^{3 / 2}-\frac{1}{18}(2 x-3)^{3 / 2}+C \text {. }\)

\(\int \frac{d x}{(\sqrt{2 x+3}+\sqrt{2 x-3})}\) = \(\frac{1}{18}(2 x+3)^{3 / 2}-\frac{1}{18}(2 x-3)^{3 / 2}+C \text {. }\)

Example 15 Evaluate:

(1) \(\int \frac{1}{(1+\tan x)} d x\)

(2) \(\int \frac{1}{(1+\cot x)} d x\)

(3) \(\int\left(\frac{1-\tan x}{1+\tan x}\right) d x\)

(4) \(\int \frac{\tan x}{(\sec x+\cos x)} d x\)

Integration Types Maths

Solution

(1) \(\int \frac{1}{(1+\tan x)} d x=\int \frac{1}{\left(1+\frac{\sin x}{\cos x}\right)} d x\)

= \(\int \frac{\cos x}{(\cos x+\sin x)} d x=\int \frac{(\cos x+\sin x)+(\cos x-\sin x)}{2(\cos x+\sin x)} d x\)

= \(\frac{1}{2} \int d x+\frac{1}{2} \int \frac{(\cos x-\sin x)}{(\cos x+\sin x)} d x\)

= \(\frac{1}{2} \int d x+\frac{1}{2} \int \frac{1}{t} d t\), where (cos x + sin x) = t and (cos x – sin x)dx = dt

= \(\frac{1}{2} x+\frac{1}{2} \log |t|+C=\frac{1}{2} x+\frac{1}{2} \log |\cos x+\sin x|+C\)

(2) \(\int \frac{1}{(1+\cot x)} d x=\int \frac{1}{\left(1+\frac{\cos x}{\sin x}\right)} d x=\int \frac{\sin x}{(\sin x+\cos x)} d x\)

= \(\int \frac{(\sin x+\cos x)-(\cos x-\sin x)}{2(\sin x+\cos x)} d x\)=\(\frac{1}{2} \int d x-\frac{1}{2} \int \frac{(\cos x-\sin x)}{(\sin x+\cos x)} d x\)

Integration Types Maths

= \(\frac{1}{2} \int d x-\frac{1}{2} \int \frac{1}{t} d t\), where sin x + cos x = t and (cos x – sin x)dx = dt

= \(\frac{1}{2} x-\frac{1}{2} \log |t|+C=\frac{1}{2} x-\frac{1}{2} \log |\sin x+\cos x|+C .\)

(3) \(\int\left(\frac{1-\tan x}{1+\tan x}\right) d x=\int \frac{\left(1-\frac{\sin x}{\cos x}\right)}{\left(1+\frac{\sin x}{\cos x}\right)} d x=\int \frac{(\cos x-\sin x)}{(\cos x+\sin x)} d x\)

= \(\int \frac{1}{t} d t\), where (cos x + sin x) = t and (cos x – sin x)dx = dt

= log|t| + C = log |(cos x + sin x)| + C.

(4) \(\int \frac{\tan x}{(\sec x+\cos x)} d x=\int \frac{\sin x}{1+\cos ^2 x} d x\)

= \(-\int \frac{1}{\left(1+t^2\right)} d t\), where cos x = t and sin x dx = -dt

= -tan-1t + C = -tan-1(cos x) + C.

Example 16 Evaluate:

(1) \(\int \tan x d x\)

(2) \(\int \cot x d x\)

(3) \(\int \sec x d x\)

Solution

(1) \(\int \tan x d x=\int \frac{\sin x}{\cos x} d x\)

= \(-\int \frac{1}{t} d t\), where cos x = t and sin x dx = -dt

= -log |t| + C = -log |cos x| + C.

∴ \(\int \tan x d x=-\log |\cos x|+C .\)

(2) \(\int \cot x d x=\int \frac{\cos x}{\sin x} d x\)

Integration Types Maths

= \(\int \frac{1}{t} d t\), where sin x = t and cos x dx = dt

= log |t| + C = log |sin x| + C.

∴ \(\int \cot x d x=\log |\sin x|+C\) .

(3) \(\int \sec x d x=\int \frac{\sec x(\sec x+\tan x)}{(\sec x+\tan x)} d x\)

[multiplying numerator and denominator by (sec x + tan x)]

= \(\int \frac{1}{t} d t\), where (sec x + tan x) = t and sec x(sec x + tan x)dx = dt

= log |t| + C = log |(sec x + tam x)] + C.

∴ \(\int \sec x d x=\log |(\sec x+\tan x)|+C .\)

Alternative form

\(\sec x+\tan x=\left(\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right)=\frac{(1+\sin x)}{\cos x}\)

Putting \(\sin x=\frac{2 \tan (x / 2)}{1+\tan ^2(x / 2)} \text { and } \cos x=\frac{1-\tan ^2(x / 2)}{1+\tan ^2(x / 2)} \text {. }\)

∴ \(\sec x+\tan x=\frac{1+\tan (x / 2)}{1-\tan (x / 2)}=\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\)

∴ \(\int \sec x d x=\log |\sec x+\tan x|+C\)

= \(\log \left|\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right|+C\)

Types of Integration 

Class 12 Maths Integration

As a consequence of the above results, the integral of trigonometric functions may be listed as given below:

(1) \(\int \sin x d x=-\cos x+C\)

(2) \(\int \cos x d x=\sin x+C\)

(3) \(\int \tan x d x=-\log |\cos x|+C\)

(4) \(\int \cot x d x=\log |\sin x|+C\)

(5) \(\int \sec x d x=\log |\sec x+\tan x|+C=\log \left|\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right|+C\)

(6) \(\int {cosec} x d x=\log |{cosec} x-\cot x|+C=\log \left|\tan \frac{x}{2}\right|+C\)

Integration by Parts Step-by-Step Guide

Example 17 Evaluate:

(1) \(\int \frac{(1+\cos x)}{(1-\cos x)} d x\)

(2) \(\int \frac{(1+\sin x)}{(1+\cos x)} d x\)

Solution

(1) \(\int \frac{(1+\cos x)}{(1-\cos x)} d x=\int \frac{2 \cos ^2(x / 2)}{2 \sin ^2(x / 2)} d x\)

= \(\int \cot ^2\left(\frac{x}{2}\right) d x=\int\left({cosec}^2 \frac{x}{2}-1\right) d x\)

= \(2 \int {cosec}^2 t d t-\int d x\), where \(\frac{x}{2}\) = t and dx = 2dt

= \(-2 \cot t-x+C=-2 \cot \left(\frac{x}{2}\right)-x+C\).

(2) \(\int\left(\frac{1+\sin x}{1+\cos x}\right) d x=\int \frac{1}{(1+\cos x)} d x+\int \frac{\sin x}{(1+\cos x)} d x\)

= \(\int \frac{1}{2 \cos ^2(x / 2)} d x+\int \frac{2 \sin (x / 2) \cos (x / 2)}{2 \cos ^2(x / 2)} d x\)

= \(\frac{1}{2} \int \sec ^2\left(\frac{x}{2}\right) d x+\int \tan \frac{x}{2} d x\)

= \(\int \sec ^2 t d t+2 \int \tan t d t\), where \(\frac{x}{2}\) = t

Class 12 Maths Integration

= tan t – 2 log|cos t| + C

= \(\tan \left(\frac{x}{2}\right)-2 \log \left|\cos \left(\frac{x}{2}\right)\right|+C\).

Example 18 Evaluate:

(1) \(\int \frac{d x}{1+\sqrt{x}}\)

(2) \(\int \frac{x+\sqrt{x+1}}{x+2} d x\)

Solution

(1) Put √x = t so that x = t² and dx = 2t dt.

∴ \(\int \frac{x+\sqrt{x+1}}{(x+2)} d x=2 \int \frac{\left(t^2-1+t\right) t}{\left(t^2+1\right)} d t\)

= \(2 \int\left(\frac{t^3+t^2-t}{t^2+1}\right) d t\)

= \(2 \int\left(t+1-\frac{2 t+1}{t^2+1}\right) d t\) [by division]

= \(2 \int\left(t+1-\frac{2 t}{t^2+1}-\frac{1}{t^2+1}\right) d t\)

= \(2 \int t d t+2 \int d t-2 \int \frac{2 t}{t^2+1} d t-2 \int \frac{1}{t^2+1} d t\)

= t² + 2t – 2 log |t²+1| – 2tan^-1t + C

= \((x+1)+2 \sqrt{x+1}-2 \log |x+2|-2 \tan ^{-1} \sqrt{x+1}+C\).

Example 19 Evaluate \(\int \sqrt{\frac{1+x}{1-x}} d x\)

Solution

Given

\(\int \sqrt{\frac{1+x}{1-x}} d x\) \(\int \sqrt{\frac{1+x}{1-x}} d x=\int \frac{\sqrt{1+x}}{\sqrt{1-x}} \times \frac{\sqrt{1+x}}{\sqrt{1+x}} d x\)

= \(\int \frac{1+x}{\sqrt{1-x^2}} d x=\int \frac{d x}{\sqrt{1-x^2}}+\int \frac{x}{\sqrt{1-x^2}} d x\)

= \(\int \frac{d x}{\sqrt{1-x^2}}-\frac{1}{2} \int \frac{1}{\sqrt{t}} d t\), where (1-x²) = t

Class 12 Maths Integration

= sin^-1x – √t + C

= \(\sin ^{-1} x-\sqrt{1-x^2}+C\).

\(\int \sqrt{\frac{1+x}{1-x}} d x\) = \(\sin ^{-1} x-\sqrt{1-x^2}+C\).

Types of Integration 

Integration Using Trigonometric Identities

When the integrand consists of trigonometric functions, we use known identities to convert it into a form which can easily be integrated. Some of the identities useful for this purpose are given below:

(1) \(2 \sin ^2\left(\frac{x}{2}\right)=(1-\cos x)\)

(2) \(2 \cos ^2\left(\frac{x}{2}\right)=(1+\cos x)\)

(3) 2 sin a cos b = sin (a+b) + sin (a-b)

(4) 2 cos a sin b = sin (a+b) – sin (a-b)

(5) 2 cos a cos b = cos (a+b) + cos (a-b)

(6) 2 sin a sin b = cos (a-b) – cos (a+b)

Class 12 Maths Integration Solved Examples

Example 1 Evaluate:

(1) \(\int \sin ^2 \frac{x}{2} d x\)

(2) \(\int \tan ^2 \frac{x}{2} d x\)

(3) \(\int \cos ^2 n x d x\)

(4) \(\int \cos ^5 x d x\)

(5) \(\int \sin ^7 x d x\)

(6) \(\int \sin ^3(2 x+1) d x\)

Solution

(1) \(\int \sin ^2 \frac{x}{2} d x=\frac{1}{2} \int 2 \sin ^2 \frac{x}{2} d x\)

= \(\frac{1}{2} \int(1-\cos x) d x=\frac{1}{2} \int d x-\frac{1}{2} \int \cos x d x\)

= \(\frac{1}{2} x-\frac{1}{2} \sin x+C\)

(2) \(\int \tan ^2 \frac{x}{2} d x=\int\left(\sec ^2 \frac{x}{2}-1\right) d x=\int \sec ^2 \frac{x}{2} d x-\int d x\)

= \(2 \int \sec ^2 t d t-\int d x\), where \(\frac{x}{2}\) = t

= \(2 \tan t-x+C=2 \tan \frac{x}{2}-x+C .\)

(3) \(\int \cos ^2 n x d x=\frac{1}{2} \int 2 \cos ^2 n x d x\)

= \(\frac{1}{2} \int(1+\cos 2 n x) d x=\frac{1}{2} \int d x+\frac{1}{2} \int \cos 2 n x d x\)

= \(\frac{x}{2}+\frac{1}{4 n} \sin 2 n x+C\)

(4) \(\int \cos ^5 x d x=\int \cos ^4 x \cdot \cos x d x\)

= \(\int\left(1-\sin ^2 x\right)^2 \cdot \cos x d x=\int\left(1-t^2\right)^2 d t\), where sin x = t

= \(\int\left(1+t^4-2 t^2\right) d t=\int d t+\int t^4 d t-2 \int t^2 d t\)

= \(t+\frac{t^5}{5}-\frac{2 t^3}{3}+C=\sin x+\frac{1}{5} \sin ^5 x-\frac{2}{3} \sin ^3 x+C\).

(5) \(\int \sin ^7 x d x=\int \sin ^6 x \cdot \sin x d x\)

Methods of Integration 

Class 12 Maths Integration

= \(\int\left(1-\cos ^2 x\right)^3 \sin x d x\)

= \(-\int\left(1-t^2\right)^3 d t\), where cos x = t

= \(\int\left(t^6-3 t^4+3 t^2-1\right) d t=\frac{t^7}{7}-\frac{3 t^5}{5}+t^3-t+C\)

= \(\frac{1}{7} \cos ^7 x-\frac{3}{5} \cos ^5 x+\cos ^3 x-\cos x+C\)

(6) \(\int \sin ^3(2 x+1) d x=\int\left\{1-\cos ^2(2 x+1)\right\} \cdot \sin (2 x+1) d x\)

= \(-\frac{1}{2} \int\left(1-t^2\right) d t\), where cos(2x+1) = t-1

= \(-\frac{1}{2} \int d t+\frac{1}{2} \int t^2 d t=-\frac{1}{2} t+\frac{1}{6} t^3+C\)

= \(-\frac{1}{2} \cos (2 x+1)+\frac{1}{6} \cos ^3(2 x+1)+C .\)

Trigonometric Integration Techniques

Example 2 Evaluate \(\int \cos m x \cos n x d x\), when (1) m ≠ n (2) m = n.

Solution

Given

\(\int \cos m x \cos n x d x\), when (1) m ≠ n (2) m = n.

(1) When m ≠ n, we have

\(\int \cos m x \cos n x d x=\frac{1}{2} \int[\cos (m+n) x+\cos (m-n) x] d x\)

= \(\frac{1}{2} \int \cos (m+n) x d x+\frac{1}{2} \int \cos (m-n) x d x\)

= \(\frac{\sin (m+n) x}{2(m+n)}+\frac{\sin (m-n) x}{2(m-n)}+C .\)

(2) When m = n, we have

\(\int \cos m x \cos n x d x=\int \cos ^2 n x d x\)

= \(\frac{1}{2} \int 2 \cos ^2 n x d x=\frac{1}{2} \int(1+\cos 2 n x) d x\)

= \(\frac{1}{2} \int d x+\frac{1}{2} \int \cos 2 n x d x=\frac{x}{2}+\frac{\sin 2 n x}{4 n}+C .\)

Example 3 Evaluate:

(1) \(\int \sin 3 x \sin 2 x d x\)

(2) \(\int \cos 3 x \sin 2 x d x\)

Class 12 Maths Integration

(3) \(\int \cos 4 x \cos x d x\)

(4) \(\int \sin ^3 x \cos ^3 x d x\)

Solution

(1) Using 2 sin a sin b = cos(a-b) – cos(a+b), we have

\(\int \sin 3 x \sin 2 x d x=\frac{1}{2} \int 2 \sin 3 x \sin 2 x d x\)

= \(\frac{1}{2} \int(\cos x-\cos 5 x) d x\)

= \(\frac{1}{2} \int \cos x d x-\frac{1}{2} \int \cos 5 x d x=\frac{1}{2} \sin x-\frac{\sin 5 x}{10}+\text { C. }\)

(2) Using 2 cos a sin b = sin (a+b) – sin (a-b), we get

\(\int \cos 3 x \sin 2 x d x=\frac{1}{2} \int 2 \cos 3 x \sin 2 x d x\)

= \(\frac{1}{2} \int(\sin 5 x-\sin x) d x\)

= \(\frac{1}{2} \int \sin 5 x d x-\frac{1}{2} \int \sin x d x\)

= \(\frac{-\cos 5 x}{10}+\frac{\cos x}{2}+C .\)

(3) Using 2 cos a cos b = cos(a+b) + cos(a-b), we get

\(\int \cos 4 x \cos x d x=\frac{1}{2} \int 2 \cos 4 x \cos x d x\)

= \(\frac{1}{2} \int(\cos 5 x+\cos 3 x) d x\)

= \(\frac{1}{2} \int \cos 5 x d x+\frac{1}{2} \int \cos 3 x d x\)

= \(\frac{\sin 5 x}{10}+\frac{\sin 3 x}{6}+C .\)

(4) \(\int \sin ^3 x \cos ^3 x d x=\int \sin ^3 x \cos ^2 x \cos x d x\)

= \(\int \sin ^3 x\left(1-\sin ^2 x\right) \cos x d x\)

= \(\int t^3\left(1-t^2\right) d t\), where sin x = t

= \(\int t^3 d t-\int t^5 d t=\frac{t^4}{4}-\frac{t^6}{6}+\mathrm{C}\)

= \(\frac{1}{4} \sin ^4 x-\frac{1}{6} \sin ^6 x+C\)

Example 4 Evaluate \(\int \cos x \cos 2 x \cos 3 x d x\)

Solution

Given

\(\int \cos x \cos 2 x \cos 3 x d x\) \(\int \cos x \cos 2 x \cos 3 x d x\)

= \(\frac{1}{2} \int(2 \cos x \cos 2 x) \cos 3 x d x\)

= \(\frac{1}{2} \int(\cos 3 x+\cos x) \cos 3 x d x=\frac{1}{2} \int\left(\cos ^2 3 x+\cos x \cos 3 x\right) d x\)

= \(\frac{1}{4} \int\left(2 \cos ^2 3 x\right) d x+\frac{1}{4} \int(2 \cos x \cos 3 x) d x\)

= \(\frac{1}{4} \int(1+\cos 6 x) d x+\frac{1}{4} \int(\cos 4 x+\cos 2 x) d x\)

= \(\frac{1}{4} \int d x+\frac{1}{4} \int \cos 6 x d x+\frac{1}{4} \int \cos 4 x d x+\frac{1}{4} \int \cos 2 x d x\)

= \(\frac{1}{4} x+\frac{1}{4} \cdot \frac{\sin 6 x}{6}+\frac{1}{4} \cdot \frac{\sin 4 x}{4}+\frac{1}{4} \cdot \frac{\sin 2 x}{2}+C\)

= \(\frac{x}{4}+\frac{\sin 6 x}{24}+\frac{\sin 4 x}{16}+\frac{\sin 2 x}{8}+C\)

\(\int \cos x \cos 2 x \cos 3 x d x\) = \(\frac{x}{4}+\frac{\sin 6 x}{24}+\frac{\sin 4 x}{16}+\frac{\sin 2 x}{8}+C\)

Methods of Integration 

Example 5 Evaluate \(\int \sec ^4 x \tan x d x .\)

Solution

Given

\(\int \sec ^4 x \tan x d x .\) \(\int \sec ^4 x \tan x d x=\int \sec ^2 x \cdot \sec ^2 x \tan x d x\)

= \(\int\left(1+\tan ^2 x\right) \sec ^2 x \tan x d x\)

= \(\int\left(1+t^2\right) t d t\), where tan x = t

= \(\int t d t+\int t^3 d t=\frac{t^2}{2}+\frac{t^4}{4}+C\)

= \(\frac{1}{2} \tan ^2 x+\frac{1}{4} \tan ^4 x+C\)

\(\int \sec ^4 x \tan x d x .\) = \(\frac{1}{2} \tan ^2 x+\frac{1}{4} \tan ^4 x+C\)

Example 6 Evaluate \(\int \sin ^4 x d x\)

Solution

Given

\(\int \sin ^4 x d x\) \(\int \sin ^4 x d x=\frac{1}{4} \int\left(2 \sin ^2 x\right)^2 d x\)

= \(\frac{1}{4} \int(1-\cos 2 x)^2 d x=\frac{1}{4} \int\left(1+\cos ^2 2 x-2 \cos 2 x\right) d x\)

= \(\frac{1}{8} \int\left(2+2 \cos ^2 2 x-4 \cos 2 x\right) d x\)

= \(\frac{1}{8} \int[2+(1+\cos 4 x)-4 \cos 2 x] d x\)

= \(\frac{3}{8} \int d x+\frac{1}{8} \int \cos 4 x d x-\frac{1}{2} \int \cos 2 x d x\)

= \(\frac{3}{8} x+\frac{\sin 4 x}{32}-\frac{\sin 2 x}{4}+C\)

\(\int \sin ^4 x d x\) = \(\frac{3}{8} x+\frac{\sin 4 x}{32}-\frac{\sin 2 x}{4}+C\)

Example 7 Evaluate \(\int \frac{\sin x}{\sin (x-\alpha)} d x \text {. }\)

Solution

Given

\(\int \frac{\sin x}{\sin (x-\alpha)} d x \text {. }\)

Put (x-α) = t so that x = (t+α) and dx = dt.

∴ \(\int \frac{\sin x}{\sin (x-\alpha)} d x=\int \frac{\sin (t+\alpha)}{\sin t} d t\)

= \(\int \frac{\sin t \cos \alpha+\cos t \sin \alpha}{\sin t} d t\)

= \(\cos \alpha \cdot \int d t+\sin \alpha \cdot \int \cot t d t\)

= cos α . t + sin α . log |sin t| + C

= (cos α)(x-α) + sin α. log |sin (x-α)| + C.

\(\int \frac{\sin x}{\sin (x-\alpha)} d x \text {. }\) = (cos α)(x-α) + sin α. log |sin (x-α)| + C.

Example 8 Evaluate:

(1) \(\int \frac{\sin 4 x}{\cos 2 x} d x\)

(2) \(\int \frac{\sin 4 x}{\sin x} d x\)

Solution

(1) \(\int \frac{\sin 4 x}{\cos 2 x} d x=\int \frac{2 \sin 2 x \cos 2 x}{\cos 2 x} d x\)

= \(2 \int \sin 2 x d x=-\cos 2 x+C .\)

(2) \(\int \frac{\sin 4 x}{\sin x} d x=\int \frac{2 \sin 2 x \cos 2 x}{\sin x} d x\)

= \(\int \frac{4 \sin x \cos x \cos 2 x}{\sin x} d x=2 \int 2 \cos x \cos 2 x d x\)

= \(2 \int(\cos 3 x+\cos x) d x=2 \int \cos 3 x d x+2 \int \cos x d x\)

= \(\frac{2 \sin 3 x}{3}+2 \sin x+C \text {. }\)

Methods of Integration 

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Partial Fraction Integration Method

Example 9 Evaluate \(\int \sqrt{1+\sin x} d x\)

Solution

Given

\(\int \sqrt{1+\sin x} d x\) \(\int \sqrt{1+\sin x} d x=\int \sqrt{\sin ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}} d x\)

= \(\int\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right) d x=\int \sin \frac{x}{2} d x+\int \cos \frac{x}{2} d x\)

= \(-2 \cos \frac{x}{2}+2 \sin \frac{x}{2}+C\)

\(\int \sqrt{1+\sin x} d x\) = \(-2 \cos \frac{x}{2}+2 \sin \frac{x}{2}+C\)

Example 10 Evaluate \(\int \frac{\sin ^2 x}{(1+\cos x)^2} d x\)

Solution

Given

\(\int \frac{\sin ^2 x}{(1+\cos x)^2} d x\) \(\int \frac{\sin ^2 x}{(1+\cos x)^2} d x=\int\left(\frac{\sin x}{1+\cos x}\right)^2 d x=\int\left[\frac{2 \sin (x / 2) \cos (x / 2)}{2 \cos ^2(x / 2)}\right]^2 d x\)

= \(\int \tan ^2(x / 2) d x=\int\left(\sec ^2 \frac{x}{2}-1\right) d x\)

= \(\int \sec ^2(x / 2) d x-\int d x=2 \tan (x / 2)-x+C .\)

\(\int \frac{\sin ^2 x}{(1+\cos x)^2} d x\) = \(\int \sec ^2(x / 2) d x-\int d x=2 \tan (x / 2)-x+C .\)

Example 11 Evaluate:

(1) \(\int \sin x \sqrt{1-\cos 2 x} d x\)

(2) \(\int \frac{\cos 2 x}{\sqrt{1+\cos 4 x}} d x\)

Solution

(1) \(\int \sin x \sqrt{1-\cos 2 x} d x\)

= \(\int \sin x \cdot \sqrt{2 \sin ^2 x} d x=\sqrt{2} \int \sin ^2 x d x=\frac{1}{\sqrt{2}} \int 2 \sin ^2 x d x\)

= \(\frac{1}{\sqrt{2}} \int(1-\cos 2 x) d x=\frac{1}{\sqrt{2}} \int d x-\frac{1}{\sqrt{2}} \int \cos 2 x d x\)

= \(\frac{1}{\sqrt{2}} x-\frac{\sin 2 x}{2 \sqrt{2}}+C .\)

(2) \(\int \frac{\cos 2 x}{\sqrt{1+\cos 4 x}} d x=\int \frac{\cos 2 x}{\sqrt{2 \cos ^2 2 x}} d x=\frac{1}{\sqrt{2}} \int d x=\frac{x}{\sqrt{2}}+C\).

Example 12 Evaluate:

(1) \(\int \frac{d x}{a \sin x+b \cos x}\)

(2) \(\int \frac{d x}{\sin x+\cos x}\)

Solution

(1) Put a = r cosθ and b = r sinθ so that

r² = (a²+b²) and θ = tan^2-1(b/a).

∴ \(\int \frac{d x}{a \sin x+b \cos x}=\int \frac{d x}{r \cos \theta \sin x+r \sin \theta \cos x}\)

= \(\frac{1}{r} \int \frac{d x}{\sin (x+\theta)}=\frac{1}{r} \cdot \int {cosec}(x+\theta) d x\)

= \(\frac{1}{r} \log \left\{\tan \left(\frac{\theta+x}{2}\right)\right\}+C\)

= \(\frac{1}{\sqrt{a^2+b^2}} \log \left[\tan \left\{\frac{1}{2} \tan ^{-1}\left(\frac{b}{a}\right)+\frac{x}{2}\right\}\right]+C.\)

(2) We can write,

Integration Examples

\(\int \frac{d x}{\sin x+\cos x}=\frac{1}{\sqrt{2}} \int \frac{d x}{\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x}\)

= \(\frac{1}{\sqrt{2}} \int \frac{d x}{\left(\cos \frac{\pi}{4} \sin x+\sin \frac{\pi}{4} \cos x\right)}\)

= \(\frac{1}{\sqrt{2}} \int \frac{d x}{\sin \left(\frac{\pi}{4}+x\right)}=\frac{1}{\sqrt{2}} \int {cosec}\left(\frac{\pi}{4}+x\right) d x\)

= \(\frac{1}{\sqrt{2}} \log \tan \left(\frac{\pi}{8}+\frac{x}{2}\right)+C .\)

Methods of Integration 

Example 13 Evaluate:

(1) \(\int \frac{d x}{4 \cos x+3 \sin x}\)

(2) \(\int \frac{d x}{(2 \sin x+3 \cos x)^2}\)

Solution

(1) Put 4 = r sin θ and 3 = r cos θ so that

r² = 25 and \(\theta=\tan ^{-1}\left(\frac{4}{3}\right)\)

∴ \(\int \frac{d x}{4 \cos x+3 \sin x}=\int \frac{d x}{r \sin \theta \cos x+r \cos \theta \sin x}\)

= \(\frac{1}{r} \int \frac{d x}{\sin (\theta+x)}=\frac{1}{r} \int {cosec}(\theta+x) d x\)

= \(\frac{1}{r} \log \left\{\tan \left(\frac{\theta+x}{2}\right)\right\}+C\)

= \(\frac{1}{5} \log \left[\tan \left\{\frac{1}{2} \tan ^{-1}\left(\frac{4}{3}\right)+\frac{x}{2}\right\}\right]+C\)

(2) Put 2 = r cos θ and 3 = r sin θ so that r² = 13 and \(\theta=\tan ^{-1}\left(\frac{3}{2}\right)\)

∴ \(\int \frac{d x}{(2 \sin x+3 \cos x)^2}=\int \frac{d x}{(r \cos \theta \sin x+r \sin \theta \cos x)^2}\)

= \(\frac{1}{r^2} \int \frac{d x}{\sin ^2(\theta+x)}=\frac{1}{13} \cdot \int {cosec}^2(\theta+x) d x\)

= \(-\frac{1}{13} \cot (\theta+x)+C\)

= \(-\frac{1}{13} \cot \left\{\tan ^{-1}\left(\frac{3}{2}\right)+x\right\}+C \text {. }\)

Example 14 Evaluate \(\int \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x\)

Solution

Given

\(\int \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x\) \(\int \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x=\int \frac{\cos ^2(x / 2)-\sin ^2(x / 2)}{\{\cos (x / 2)+\sin (x / 2)\}^3} d x\)

= \(\int \frac{\cos (x / 2)-\sin (x / 2)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2} d x=2 \int \frac{1}{t^2} d t\), where t = cos\(\frac{x}{2}\) + sin\(\frac{x}{2}\)

= \(\frac{-2}{t}+C=\frac{-2}{\cos (x / 2)+\sin (x / 2)}+C\)

\(\int \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x\) = \(\frac{-2}{t}+C=\frac{-2}{\cos (x / 2)+\sin (x / 2)}+C\)

Example 15 Evaluate \(\int \frac{d x}{\sqrt{1-\sin x}}\)

Solution

Given 

\(\int \frac{d x}{\sqrt{1-\sin x}}\) \(\int \frac{d x}{\sqrt{1-\sin x}}=\int \frac{d x}{\left[\sin ^2(x / 2)+\cos ^2(x / 2)-2 \sin \frac{x}{2} \cos \frac{x}{2}\right]^{1 / 2}}\)

= \(\int \frac{d x}{\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)}=\frac{1}{\sqrt{2}} \int \frac{d x}{\left(\frac{1}{\sqrt{2}} \cdot \sin \frac{x}{2}-\cos \frac{x}{2} \cdot \frac{1}{\sqrt{2}}\right)}\)

= \(\frac{1}{\sqrt{2}} \cdot \int \frac{d x}{\left(\sin \frac{x}{2} \cos \frac{\pi}{4}-\cos \frac{x}{2} \sin \frac{\pi}{4}\right)}\)

= \(\frac{1}{\sqrt{2}} \int {cosec}\left(\frac{x}{2}-\frac{\pi}{4}\right) d x=\frac{1}{\sqrt{2}} 2 \cdot \log \left[\tan \left(\frac{x}{4}-\frac{\pi}{8}\right)\right]+C\)

= \(\sqrt{2} \log \tan \left(\frac{x}{4}-\frac{\pi}{8}\right)+C\).

\(\int \frac{d x}{\sqrt{1-\sin x}}\) = \(\sqrt{2} \log \tan \left(\frac{x}{4}-\frac{\pi}{8}\right)+C\).

Integration Examples

Example 16 Evaluate \(\int \frac{\sin x}{\sqrt{1+\sin x}} d x\)

Solution

Given

\(\int \frac{\sin x}{\sqrt{1+\sin x}} d x\) \(\int \frac{\sin x}{\sqrt{1+\sin x}} d x=\int \frac{(1+\sin x)-1}{\sqrt{1+\sin x}} d x\)

= \(\int \sqrt{1+\sin x} d x-\int \frac{d x}{\sqrt{1+\sin x}}\)

= \(\int \sqrt{\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}} d x-\int \frac{d x}{\sqrt{\cos ^2(x / 2)+\sin ^2(x / 2)+2 \sin (x / 2) \cos (x / 2)}}\)

= \(\int[\cos (x / 2)+\sin (x / 2)] d x-\int \frac{d x}{[\cos (x / 2)+\sin (x / 2)]}\)

= \(\left(2 \sin \frac{x}{2}-2 \cos \frac{x}{2}\right)-\frac{1}{\sqrt{2}} \cdot \int \frac{d x}{\frac{1}{\sqrt{2}} \cos \frac{x}{2}+\frac{1}{\sqrt{2}} \sin \frac{x}{2}}\)

= \(\left(2 \sin \frac{x}{2}-2 \cos \frac{x}{2}\right)-\frac{1}{\sqrt{2}} \cdot \int \frac{d x}{\sin \left(\frac{x}{2}+\frac{\pi}{4}\right)}\)

= \(\left(2 \sin \frac{x}{2}-2 \cos \frac{x}{2}\right)-\frac{1}{\sqrt{2}} \int {cosec}\left(\frac{x}{2}+\frac{\pi}{4}\right) d x\)

= \(2\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)-\frac{1}{\sqrt{2}} \times 2 \log \left|\tan \left(\frac{x}{4}+\frac{\pi}{8}\right)\right|+C\)

= \(2\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)-\sqrt{2} \log \left|\tan \left(\frac{x}{4}+\frac{\pi}{8}\right)\right|+C .\)

\(\int \frac{\sin x}{\sqrt{1+\sin x}} d x\) = \(2\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)-\sqrt{2} \log \left|\tan \left(\frac{x}{4}+\frac{\pi}{8}\right)\right|+C .\)

Integration Examples 

Integration By Parts

Theorem If u and v are two functions of x then

\(\int(u v) d x=\left[u \cdot \int v d x\right]-\int\left\{\frac{d u}{d x} \cdot \int v d x\right\} d x\)

Proof

For any two functions f₁(x) and f₁(x), we have

\(\frac{d}{d x}\left[f_1(x) \cdot f_2(x)\right]=f_1(x) \cdot f_2^{\prime}(x)+f_2(x) \cdot f_1^{\prime}(x) .\)

∴ \(\int\left\{f_1(x) \cdot f_2^{\prime}(x)+f_2(x) \cdot f_1^{\prime}(x)\right\} d x=f_1(x) \cdot f_2(x)\)

or \(\int f_1(x) \cdot f_2^{\prime}(x) d x+\int f_2(x) \cdot f_1^{\prime}(x) d x=f_1(x) \cdot f_2(x)\)

or \(\int f_1(x) \cdot f_2^{\prime}(x) d x=f_1(x) \cdot f_2(x)-\int f_2(x) \cdot f_1^{\prime}(x) d x \text {. }\)

Let f₁(x) = u and f₂‘(x) = v so that f₂(x) = \(\int v d x\)

Integration Examples

∴ \(\int(u v) d x=u \cdot \int v d x-\int\left\{\frac{d u}{d x} \cdot \int v d x\right\} d x\)

We can express this result as given below:

Integral of product of two functions

= (1st function) x (integral of 2nd) – \(\left.\int\{(\text { derivative of } 1 \text { st }) \times \text { (integral of } 2 \text { nd })\right\} d x\)

Remarks

(1) If the integrand is of the form f(x)xⁿ, we consider xⁿ as the first function and f(x) as the second function.

(2) If the integrated contains a logarithmic or an inverse trigonometric function, we take it as the first function. In all such cases, if the second function is not given, we take it as 1.

Example 1 Evaluate:

(1) \(\int x \sec ^2 x d x\)

(2) \(\int x \sin 2 x d x\)

Solution

(1) Integrating by parts, taking x as the first function, we get

\(\int x \sec ^2 x d x=x \cdot \int \sec ^2 x d x-\int\left\{\frac{d}{d x}(x) \cdot \int \sec ^2 x d x\right\} d x\)

= \(x \tan x-\int 1 \cdot \tan x d x=x \tan x+\log |\cos x|+C\).

(2) Integrating by parts, taking x as the first function, we get

\(\int x \sin 2 x d x=x \cdot \int \sin 2 x d x-\int\left\{\frac{d}{d x}(x) \cdot \int \sin 2 x d x\right\} d x\)

= \(x \cdot\left(\frac{-\cos 2 x}{2}\right)-\int 1 \cdot\left(\frac{-\cos 2 x}{2}\right) d x\)

= \(\frac{-x \cos 2 x}{2}+\frac{1}{2} \int \cos 2 x d x\)

= \(\frac{-x \cos 2 x}{2}+\frac{1}{2} \cdot \frac{\sin 2 x}{2}+C\)

= \(\frac{-x \cos 2 x}{2}+\frac{1}{4} \sin 2 x+C .\)

Applications of Integration in Real Life

Example 2 Evaluate \(\int x^n \log x d x\)

Solution

Given

\(\int x^n \log x d x\)

Integrating by parts, taking log x as the first function, we get

\(\int x^n \log x=(\log x) \cdot \int x^n d x-\int\left\{\frac{d}{d x}(\log x) \cdot \int x^n d x\right\} d x\)

= \((\log x) \cdot \frac{x^{n+1}}{(n+1)}-\int \frac{1}{x} \cdot \frac{x^{n+1}}{(n+1)} d x\)

= \(\frac{x^{n+1} \log x}{(n+1)}-\frac{1}{(n+1)} \int x^n d x\)

= \(\frac{x^{n+1} \log x}{(n+1)}-\frac{x^{n+1}}{(n+1)^2}+C\).

\(\int x^n \log x d x\) = \(\frac{x^{n+1} \log x}{(n+1)}-\frac{x^{n+1}}{(n+1)^2}+C\).

Example 3 Evaluate \(\int x^2 \sin x d x\)

Solution

Given:

\(\int x^2 \sin x d x\)

Integrating by parts, taking x² as the first function, we get

\(\int x^2 \sin x d x=x^2 \int \sin x d x-\int\left[\frac{d}{d x}\left(x^2\right) \cdot \int \sin x d x\right] d x\)

= \(x^2(-\cos x)-\int 2 x(-\cos x) d x\)

= \(-x^2 \cos x+2 \int x \cos x d x\)

= \(-x^2 \cos x+2\left[x(\sin x)-\int\left\{\frac{d}{d x}(x) \cdot \int \cos x d x\right\} d x\right]\) [integrating x cos x by parts]

= \(-x^2 \cos x+2\left[x \sin x-\int \sin x d x\right]\)

= \(-x^2 \cos x+2[x \sin x+\cos x]+C\)

\(\int x^2 \sin x d x\) = \(-x^2 \cos x+2[x \sin x+\cos x]+C\)

Integration Examples 

Example 4 Evaluate \(\int x \cos ^2 x d x\)

Solution

Given:

\(\int x \cos ^2 x d x\) \(\int x \cos ^2 x d x=\int x\left(\frac{1+\cos 2 x}{2}\right) d x=\frac{1}{2} \int x d x+\frac{1}{2} \int x \cos 2 x d x\)

= \(\frac{x^2}{4}+\frac{1}{2} \cdot\left[x \cdot \int \cos 2 x d x-\int\left\{\frac{d}{d x}(x) \cdot \int \cos 2 x d x\right\} d x\right]\) [integrating x cos 2x by parts]

= \(\frac{x^2}{4}+\frac{1}{2}\left[\frac{x \sin 2 x}{2}-\int \frac{\sin 2 x}{2} d x\right]\)

= \(\frac{x^2}{4}+\frac{x \sin 2 x}{4}-\frac{1}{4} \cdot \frac{(-\cos 2 x)}{2}+C\)

= \(\frac{x^2}{4}+\frac{x \sin 2 x}{4}+\frac{\cos 2 x}{8}+C .\)

\(\int x \cos ^2 x d x\) = \(\frac{x^2}{4}+\frac{x \sin 2 x}{4}+\frac{\cos 2 x}{8}+C .\)

Example 5 Evaluate \(\int \log x d x\)

Solution

Given:

\(\int \log x d x\)

Integrating by parts, taking log x as the first function and 1 as the second function, we get

\(\int \log x d x=\int(\log x \cdot 1) d x\)

= \((\log x) \cdot \int 1 d x-\int\left\{\frac{d}{d x}(\log x) \cdot \int 1 d x\right\} d x\)

= \((\log x) \cdot x-\int\left(\frac{1}{x} \cdot x\right) d x=x \log x-\int d x\)

= \(x log x – x + C = x(log x – 1) + C\).

\(\int \log x d x\) = \(x log x – x + C = x(log x – 1) + C\).

Example 6 Evaluate \(\int \log \left(1+x^2\right) d x\)

Solution

Given:

\(\int \log \left(1+x^2\right) d x\)

Integrating by parts, taking log(1+x²) as the first function and 1 as the second function, we get

= \(\int \log \left(1+x^2\right) d x=\int\left\{\log \left(1+x^2\right) \cdot 1\right\} d x\)

= \(\log \left(1+x^2\right) \cdot \int 1 d x-\int\left[\frac{d}{d x}\left\{\log \left(1+x^2\right)\right\} \cdot \int 1 d x\right] d x\)

= \(\log \left(1+x^2\right) \cdot x-\int \frac{2 x}{\left(1+x^2\right)} \cdot x d x\)

= \(x \log \left(1+x^2\right)-2 \int \frac{x^2}{\left(1+x^2\right)} d x\)

= \(x \log \left(1+x^2\right)-2 \int\left(1-\frac{1}{1+x^2}\right) d x\)

= \(x \log \left(1+x^2\right)-2 \int d x+2 \int \frac{d x}{\left(1+x^2\right)}\)

= \(x log(1+x2) – 2x + 2 tan-1x + C\).

\(\int \log \left(1+x^2\right) d x\) = \(x log(1+x2) – 2x + 2 tan-1x + C\).

Example 7 Evaluate \(\int(\log x)^2 d x\)

Solution

Given

\(\int(\log x)^2 d x\)

Integrating by parts, taking (log x)ⁿ as the first function and 1 as the second function, we get

\(\int(\log x)^2 d x=\int\left\{(\log x)^2 \cdot 1\right\} d x\)

= \((\log x)^2 \cdot \int 1 d x-\int\left\{\frac{d}{d x}(\log x)^2 \cdot \int 1 d x\right\} d x\)

= \(x(\log x)^2-\int\left(\frac{2 \log x}{x} \cdot x\right) d x\)

= \(x(\log x)^2-2 \int(\log x \cdot 1) d x\)

= \(x(\log x)^2-2\left[(\log x) \int d x-\int\left\{\frac{d}{d x}(\log x) \cdot \int d x\right\} d x\right]\)

= \(x(\log x)^2-2\left[x \log x-\int \frac{1}{x} \cdot x d x\right]\)

= x(log x)2 – 2x log x + 2x + C.

\(\int(\log x)^2 d x\) = x(log x)2 – 2x log x + 2x + C.

Integration Examples 

Example 8 Evaluate \(\int \frac{\log x}{x^2} d x .\)

Solution

Given

\(\int \frac{\log x}{x^2} d x .\)

Integrating by parts, taking log x as the first function and \(\frac{1}{x^2}\) as the second function, we get

\(\int \frac{\log x}{x^2} d x=\int(\log x) \cdot \frac{1}{x^2} d x\)

= \((\log x) \cdot \int \frac{1}{x^2} d x-\int\left\{\frac{d}{d x}(\log x) \cdot \int \frac{1}{x^2} d x\right\} d x\)

= \((\log x)\left(-\frac{1}{x}\right)-\int \frac{1}{x} \cdot\left(-\frac{1}{x}\right) d x=-\frac{\log x}{x}+\int \frac{1}{x^2} d x\)

= \(-\frac{\log x}{x}-\frac{1}{x}+C=\frac{-(\log x+1)}{x}+C\).

\(\int \frac{\log x}{x^2} d x .\) = \(-\frac{\log x}{x}-\frac{1}{x}+C=\frac{-(\log x+1)}{x}+C\).

Example 9 Evaluate \(\int e^{2 x} \sin x d x .\)

Solution

Given

\(\int e^{2 x} \sin x d x .\)

Integrating by parts, we get

\(\int e^{2 x} \sin x d x=\left(e^{2 x} \cdot \int \sin x d x\right)-\int\left\{\frac{d}{d x}\left(e^{2 x}\right) \cdot \int \sin x d x\right\} d x\)

= \(e^{2 x} \cdot(-\cos x)-2 \int e^{2 x}(-\cos x) d x\)

= \(-e^{2 x} \cos x+2 \int e^{2 x} \cos x d x\)

= \(-e^{2 x} \cos x+2\left[\left(e^{2 x} \cdot \int \cos x d x\right)-\int\left\{\frac{d}{d x}\left(e^{2 x}\right) \cdot \int \cos x d x\right\} d x\right]\)

[integrating e^2x cos x by parts]

= \(-e^{2 x} \cos x+2 e^{2 x} \sin x-4 \int e^{2 x} \sin x d x+C .\)

∴ \(5 \int e^{2 x} \sin x d x=e^{2 x}(2 \sin x-\cos x)+C\)

or \(\int e^{2 x} \sin x d x=\frac{1}{5} e^{2 x}(2 \sin x-\cos x)+\mathrm{C}\).

\(\int e^{2 x} \sin x d x .\) \(\int e^{2 x} \sin x d x=\frac{1}{5} e^{2 x}(2 \sin x-\cos x)+\mathrm{C}\).

Example 10 Evaluate \(\int\left(\frac{x-\sin x}{1-\cos x}\right) d x\)

Solution

Given:

\(\int\left(\frac{x-\sin x}{1-\cos x}\right) d x\) \(\int\left(\frac{x-\sin x}{1-\cos x}\right) d x=\int \frac{x}{(1-\cos x)} d x-\int \frac{\sin x}{(1-\cos x)} d x\)

= \(\int \frac{x}{2 \sin ^2(x / 2)} d x-\int \frac{2 \sin (x / 2) \cos (x / 2)}{2 \sin ^2(x / 2)} d x\)

= \(\frac{1}{2} \int x {cosec}^2(x / 2) d x-\int \cot (x / 2) d x\)

= \(\frac{1}{2}\left[x \cdot \int{cosec}^2(x / 2) d x-\int\left\{\frac{d}{d x}(x) \cdot \int {cosec}^2(x / 2) d x\right\} d x\right]-\int \cot (x / 2) d x\) [integrating by parts]

= \(\frac{1}{2}\left[x \cdot\left(-2 \cot \frac{x}{2}\right)-\int\left[1 \cdot\left(-2 \cot \frac{x}{2}\right)\right] d x-\int \cot (x / 2) d x+C\right.\).

= \(-x \cot \frac{x}{2}+\int \cot \frac{x}{2} d x-\int \cot \frac{x}{2} d x+C=-x \cot \frac{x}{2}+C .\)

\(\int\left(\frac{x-\sin x}{1-\cos x}\right) d x\) = \(-x \cot \frac{x}{2}+\int \cot \frac{x}{2} d x-\int \cot \frac{x}{2} d x+C=-x \cot \frac{x}{2}+C .\)

Example 11 Evaluate \(\int x \tan ^{-1} x d x .\)

Solution

Given

\(\int x \tan ^{-1} x d x .\)

Integrating by parts, taking tan^-1x as the first function, we get

\(\int x \tan ^{-1} x d x=\left(\tan ^{-1} x\right) \cdot \int x d x-\int\left\{\frac{d}{d x}\left(\tan ^{-1} x\right) \cdot \int x d x\right\} d x\)

= \(\left(\tan ^{-1} x\right) \cdot \frac{x^2}{2}-\int \frac{1}{\left(1+x^2\right)} \cdot \frac{x^2}{2} d x\)

= \(\frac{x^2 \tan ^{-1} x}{2}-\frac{1}{2} \int\left(1-\frac{1}{1+x^2}\right) d x\) [on dividing x² by 1 + x²]

= \(\frac{x^2 \tan ^{-1} x}{2}-\frac{1}{2} \int d x+\frac{1}{2} \int \frac{1}{\left(1+x^2\right)} d x\)

= \(\frac{x^2 \tan ^{-1} x}{2}-\frac{x}{2}+\frac{1}{2} \tan ^{-1} x+C\)

= \(\frac{1}{2}\left(1+x^2\right) \tan ^{-1} x-\frac{1}{2} x+C\).

\(\int x \tan ^{-1} x d x .\) = \(\frac{1}{2}\left(1+x^2\right) \tan ^{-1} x-\frac{1}{2} x+C\).

Integration Techniques 

Example 12 Evaluate \(\int x^2 \sin ^{-1} x d x\)

Solution

Given

\(\int x^2 \sin ^{-1} x d x\)

Integrating by parts, taking sin^-1x as the first function we get

\(\int x^2 \sin ^{-1} x d x=\left(\sin ^{-1} x\right) \cdot \frac{x^3}{3}-\int \frac{1}{\sqrt{1-x^2}} \cdot \frac{x^3}{3} d x\)

= \(\frac{x^3 \sin ^{-1} x}{3}-\frac{1}{3} \int \frac{x^3}{\sqrt{1-x^2}} d x\)

= \(\frac{x^3 \sin ^{-1} x}{3}-\frac{1}{3} \int \frac{x \cdot x^2}{\sqrt{1-x^2}} d x\)

= \(\frac{x^3 \sin ^{-1} x}{3}+\frac{1}{3} \int \frac{t\left(1-t^2\right)}{t} d t\), where (1-x²) = t²

= \(\frac{x^3 \sin ^{-1} x}{3}+\frac{1}{3} \int d t-\frac{1}{3} \int t^2 d t\)

= \(\frac{x^3 \sin ^{-1} x}{3}+\frac{1}{3} t-\frac{1}{9} t^3+C\)

= \(\frac{x^3 \sin ^{-1} x}{3}+\frac{1}{3} \sqrt{1-x^2}-\frac{1}{9}\left(1-x^2\right)^{3 / 2}+C\).

\(\int x^2 \sin ^{-1} x d x\) = \(\frac{x^3 \sin ^{-1} x}{3}+\frac{1}{3} \sqrt{1-x^2}-\frac{1}{9}\left(1-x^2\right)^{3 / 2}+C\).

Example 13 Evaluate:

(1) \(\int \cos ^{-1} x d x\)

(2) \(\int \tan ^{-1} x d x\)

(3) \(\int \sec ^{-1} x d x\)

Solution

(1) Put cos^-1x = t so that x = cos t and dx = -sin t dt.

∴ \(\int \cos ^{-1} x d x=-\int t \sin t d t\)

= \(-\left[t \cdot(-\cos t)-\int 1 \cdot(-\cos t) d t\right]\) [integrating by parts]

= \(t \cos t-\int \cos t d t=t \cos t-\sin t+C\)

= \(x \cos ^{-1} x-\sqrt{1-x^2}+C\)

[∵ cos t = x ⇒ \(\sin t=\sqrt{1-x^2}\)].

(2) Put tan^-1x = t so that x = tan t and dx = sec²t dt.

∴ \(\int \tan ^{-1} x d x=\int t \sec ^2 t d t\)

= \(t \cdot \tan t-\int 1 \cdot \tan t d t\) [integrating by parts]

= t . tan t + log |cos t| + C

= \(\left(\tan ^{-1} x\right) \cdot x+\log \left|\frac{1}{\sqrt{1+x^2}}\right|+C\)

[∵ tan t = x ⇒ \(\cos t=\frac{1}{\sqrt{1+x^2}}\)]

= \(x\left(\tan ^{-1} x\right)-\frac{1}{2} \log \left|1+x^2\right|+C\).

(3) Put sec^-1x = t so that x = sec t and dx = sec t tan t dt.

∴ \(\int \sec ^{-1} x d x=\int t(\sec t \tan t) d t\)

= \(t(\sec t)-\int 1 \cdot \sec t d t\) [integrating by parts]

= t (sec t) – log |sec t + tan t| + C

= \(t(\sec t)-\log \left|\sec t+\sqrt{\sec ^2 t-1}\right|+C\)

= \(x\left(\sec ^{-1} x\right)-\log \left|x+\sqrt{x^2-1}\right|+C\)

Example 14 Evaluate \(\int\left(\sin ^{-1} x\right)^2 d x\)

Solution

Given

\(\int\left(\sin ^{-1} x\right)^2 d x\)

Putting x = sin t and dx = cos t dt, we get

\(\int\left(\sin ^{-1} x\right)^2 d x=\int t^2 \cos t d t\)

= \(t^2 \cdot(\sin t)-\int 2 t(\sin t) d t\) [integrating by parts]

= \(t^2 \sin t-2\left[t(-\cos t)-\int 1 \cdot(-\cos t) d t\right]\) [integrating t(sin t) by parts]

= t2 sin t + 2t cos t – 2 sin t + C

= \(x\left(\sin ^{-1} x\right)^2+2\left(\sin ^{-1} x\right) \sqrt{1-x^2}-2 x+C .\)

\(\int\left(\sin ^{-1} x\right)^2 d x\) = \(x\left(\sin ^{-1} x\right)^2+2\left(\sin ^{-1} x\right) \sqrt{1-x^2}-2 x+C .\)

Step-by-Step Solutions to Integration Problems

Example 15 Evaluate \(\int \frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}} d x\)

Solution

Given

\(\int \frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}} d x\)

Put x = sin t so that dx = cos t dt and t = sin^-1x.

∴ \(\int \frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}} d x=\int \frac{t \cos t}{\left(1-\sin ^2 t\right)^{3 / 2}} d t=\int \frac{t \cos t}{\cos ^3 t} d t\)

= \(\int t \sec ^2 t d t\)

= \(t \cdot(\tan t)-\int 1 \cdot \tan t d t\) [integrating by parts]

= t . (tan t) + log |cos t| + C

= \(\left(\sin ^{-1} x\right) \cdot \frac{x}{\sqrt{1-x^2}}+\log \left|\sqrt{1-x^2}\right|+C\)

[∵ \(\cos t=\sqrt{1-x^2} \text { and } \tan t=\frac{x}{\sqrt{1-x^2}}\)]

= \(\frac{x\left(\sin ^{-1} x\right)}{\sqrt{1-x^2}}+\frac{1}{2} \log \left|\left(1-x^2\right)\right|+C .\)

\(\int \frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}} d x\) = \(\frac{x\left(\sin ^{-1} x\right)}{\sqrt{1-x^2}}+\frac{1}{2} \log \left|\left(1-x^2\right)\right|+C .\)

Integration Techniques 

Example 16 Evaluate \(\int \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x\)

Solution

Given

\(\int \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x\)

Put x = tan t so that dx = sec²t dt.

∴ \(\int \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x=\int \frac{(\tan t) t}{\left(1+\tan ^2 t\right)^{3 / 2}} \cdot \sec ^2 t d t\)

= \(\int \frac{(\tan t) t}{\sec t} d t=\int t \sin t d t\)

= \(t(-\cos t)-\int 1 \cdot(-\cos t) d t\) [integrating by parts]

= \(-t \cos t+\sin t+C=\frac{-\tan ^{-1} x}{\sqrt{1+x^2}}+\frac{x}{\sqrt{1+x^2}}+C\)

[∵ \(\sin t=\frac{x}{\sqrt{1+x^2}} \text { and } \cos t=\frac{1}{\sqrt{1+x^2}}\)].

\(\int \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x\)  \(\sin t=\frac{x}{\sqrt{1+x^2}} \text { and } \cos t=\frac{1}{\sqrt{1+x^2}}\)

Example 17 Evaluate:

(1) \(\int \sin ^{-1}\left(3 x-4 x^3\right) d x\)

(2) \(\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x\)

(3) \(\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x\)

(4) \(\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x\)

Solution

(1) Put x = sin t so that dx = cos t dt.

∴ \(\int \sin ^{-1}\left(3 x-4 x^3\right) d x=\int \sin ^{-1}\left(3 \sin t-4 \sin ^3 t\right) \cos t d t\)

= \(\int \sin ^{-1}(\sin 3 t) \cos t d t\)

= \(3 \int t \cos t d t\)

= \(3\left[t(\sin t)-\int 1 \cdot \sin t d t\right]\) [integrating by parts]

= 3t sin t + 3 cos t + C

= \(3 x\left(\sin ^{-1} x\right)+3 \sqrt{1-x^2}+C .\)

(2) Put x = tan t so that dx = sec²t dt.

∴ \(\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=\int \sin ^{-1}\left(\frac{2 \tan t}{1+\tan ^2 t}\right) \sec ^2 t d t\)

= \(\int \sin ^{-1}(\sin 2 t) \sec ^2 t d t=2 \int t \cdot \sec ^2 t d t\)

= \(2\left[t \cdot \tan t-\int 1 \cdot \tan t d t\right]\)

= 2t . tan t + 2 log |cos t| + C

= \(2 x\left(\tan ^{-1} x\right)+2 \log \left|\frac{1}{\sqrt{1+x^2}}\right|+C\)

= \(2 x\left(\tan ^{-1} x\right)+2 \cdot\left(-\frac{1}{2}\right) \log \left|1+x^2\right|+C\)

= 2x (tan^-1x) – log |1+x2| + C.

(3) Put x = cos t so that dx = -sin t dt.

∴ \(\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x=\int \tan ^{-1} \sqrt{\frac{1-\cos t}{1+\cos t}}(-\sin t) d t\)

= \(\int \tan ^{-1} \sqrt{\frac{2 \sin ^2(t / 2)}{2 \cos ^2(t / 2)}}(-\sin t) d t\)

= \(\int\left[\tan ^{-1}\left(\tan \frac{t}{2}\right)\right](-\sin t) d t=-\frac{1}{2} \int t(\sin t) d t\)

= \(-\frac{1}{2}\left[t(-\cos t)-\int 1 \cdot(-\cos t) d t\right]\) [integrating by parts]

= \(\frac{1}{2} t \cdot \cos t-\frac{1}{2} \sin t+C=\frac{1}{2} x\left(\cos ^{-1} x\right)-\frac{1}{2} \sqrt{1-x^2}+C .\)

(4) Put x = a tan²t so that dx = (2asec²t tan t)dt.

∴ \(\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x=\int \sin ^{-1}\left\{\sqrt{\frac{a \tan ^2 t}{a\left(1+\tan ^2 t\right)}}\right\} 2 a \sec ^2 t \tan t d t\)

= \(2 a \int t\left(\sec ^2 t \cdot \tan t\right) d t\)

= \(2 a\left[t \cdot \frac{1}{2} \tan ^2 t-\int 1 \cdot \frac{1}{2} \tan ^2 t d t\right]\)

[integrating by parts and using \(\int \sec ^2 t \tan t d t=\frac{1}{2} \tan ^2 t\)]

= \(a t\left(\tan ^2 t\right)-a \int\left(\sec ^2 t-1\right) d t\)

= \(a t\left(\tan ^2 t\right)-a \int \sec ^2 t d t+a \int d t\)

= at(tan2t) – a tan t + at + C

= \(a\left(\tan ^{-1} \sqrt{\frac{x}{a}}\right) \cdot\left(\frac{x}{a}\right)-a \cdot \sqrt{\frac{x}{a}}+a \tan ^{-1} \sqrt{\frac{x}{a}}+C\)

= \(x \tan ^{-1} \sqrt{\frac{x}{a}}-\sqrt{a x}+a \tan ^{-1} \sqrt{\frac{x}{a}}+C .\)

Example 18 Evaluate \(\int x \cos ^3 x \sin x d x\)

Solution

Given

\(\int x \cos ^3 x \sin x d x\)

Take x as the first function and (cos³x sin x) as the second.

Putting cos x = t, we can evaluate \(\int \cos ^3 x \sin x d x \text { as }-\frac{1}{4} \cos ^4 x \text {. }\)

So, integrating by parts, we get

\(\int x \cos ^3 x \sin x d x=x \cdot\left(\frac{-1}{4} \cos ^4 x\right)-\int 1 \cdot\left(-\frac{1}{4}\right) \cos ^4 x d x\)

= \(-\frac{x}{4} \cos ^4 x+\frac{1}{4} \int\left(\frac{1+\cos 2 x}{2}\right)^2 d x\)

= \(-\frac{x \cos ^4 x}{4}+\frac{1}{4} \int\left(\frac{1}{4}+\frac{\cos ^2 2 x}{4}+\cos 2 x\right) d x\)

= \(-\frac{x \cos ^4 x}{4}+\frac{1}{16} \int d x+\frac{1}{4} \int \cos 2 x d x+\frac{1}{32} \int 2 \cos ^2 2 x d x\)

= \(-\frac{x \cos ^4 x}{4}+\frac{x}{16}+\frac{\sin 2 x}{8}+\frac{1}{32} \int(1+\cos 4 x) d x+C\)

= \(-\frac{x \cos ^4 x}{4}+\frac{x}{16}+\frac{\sin 2 x}{8}+\frac{1}{32} \int d x+\frac{1}{32} \int \cos 4 x d x\)

= \(-\frac{x \cos ^4 x}{4}+\frac{3 x}{32}+\frac{\sin 2 x}{8}+\frac{\sin 4 x}{128}+C .\)

\(\int x \cos ^3 x \sin x d x\) = \(-\frac{x \cos ^4 x}{4}+\frac{3 x}{32}+\frac{\sin 2 x}{8}+\frac{\sin 4 x}{128}+C .\)

Example 19 Evaluate \(\int \sin (\log x) d x\)

Solution

Given

\(\int \sin (\log x) d x\)

Put log x = t so that x = e^t and \frac{1}{x}dx = dt or dx = e^tdt.

∴ \(\int \sin (\log x) d x=\int e^t \sin t d t\) ..(1)

Now, \(\int e^t \sin t d t=e^t(-\cos t)-\int e^t \cdot(-\cos t) d t\) [integrating by parts]

= \(-e^t \cos t+\int e^t \cos t d t\)

= \(-e^t \cos t+\left[e^t \sin t-\int e^t \sin t d t\right]\) [integratingn et cos t by parts]

= \(-e^t \cos t+e^t \sin t-\int e^t \sin t d t\)

∴ \(2 \int e^t \sin t d t=-e^t \cos t+e^t \sin t\)

or \(\int e^t \sin t d t=\frac{1}{2}\left(-e^t \cos t+e^t \sin t\right)+C\).

Putting this value in (1), we get

\(\int \sin (\log x) d x=\int e^t \sin t d t\)

= \(\frac{1}{2}\left(-e^t \cos t+e^t \sin t\right)+C\)

= \(\frac{1}{2}[-x \cos (\log x)+x \sin (\log x)]+C\)

= \(-\frac{1}{2} x \cos (\log x)+\frac{1}{2} x \sin (\log x)+C\).

\(\int \sin (\log x) d x\) = \(-\frac{1}{2} x \cos (\log x)+\frac{1}{2} x \sin (\log x)+C\).

Integration Techniques 

Example 20 Evaluate \(\int \sin \sqrt{x} d x\)

Solution

Given

\(\int \sin \sqrt{x} d x\)

Put √x = t so that \(\frac{1}{2 \sqrt{x}} d x = dt\) or dx = 2t dt.

∴ \(\int \sin \sqrt{x} d x=2 \int t \sin t d t=2\left[t(-\cos t)-\int 1 \cdot(-\cos t) d t\right]\) [integrating t sin t by parts]

= -2t cos t + 2 sin t + C

= \(-2 \sqrt{x} \cos \sqrt{x}+2 \sin \sqrt{x}+C\)

\(\int \sin \sqrt{x} d x\) = \(-2 \sqrt{x} \cos \sqrt{x}+2 \sin \sqrt{x}+C\)

Example 21 Evaluate \(\int \sec ^3 x d x\)

Solution

Given

\(\int \sec ^3 x d x\) \(\int \sec ^3 x d x=\int \sec x \cdot \sec ^2 x d x\)

= \(\sec x \cdot(\tan x)-\int \sec x \tan x(\tan x) d x\) [integrating by parts]

= \(\sec x \tan x-\int \sec x\left(\sec ^2 x-1\right) d x\)

= \(\sec x \tan x-\int \sec ^3 x d x+\int \sec x d x\)

∴ \(2 \int \sec ^3 x d x=\sec x \tan x+\log \left|\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right|+C\)

or \(\int \sec ^3 x d x=\frac{1}{2} \sec x \tan x+\frac{1}{2} \log \left|\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right|+C^{\prime} .\)

Example 22 Evaluate \(\int \tan ^{-1} \sqrt{x} d x\)

Solution

Given

\(\int \tan ^{-1} \sqrt{x} d x\)

Put √x = t so that \(\frac{1}{2 \sqrt{x}} d x=d t\) or dx = 2t dt.

∴ \(\int \tan ^{-1} \sqrt{x} d x=2 \int t\left(\tan ^{-1} t\right) d t\)

= \(2\left[\left(\tan ^{-1} t\right) \cdot \frac{t^2}{2}-\int\left\{\frac{1}{\left(1+t^2\right)} \cdot \frac{t^2}{2}\right\} d t\right]+C\)

= \(t^2\left(\tan ^{-1} t\right)-\int \frac{t^2}{\left(1+t^2\right)} d t+C\)

= \(t^2\left(\tan ^{-1} t\right)-\int \frac{\left[\left(1+t^2\right)-1\right]}{\left(1+t^2\right)} d t+C\)

= \(t^2\left(\tan ^{-1} t\right)-\int d t+\int \frac{1}{\left(1+t^2\right)} d t+C\)

= \(t2 (tan-1t) – t + tan-1t + C = (t2+1)tan-1t – t + C\)

= \((x+1) \tan ^{-1} \sqrt{x}-\sqrt{x}+C .\)

\(\int \tan ^{-1} \sqrt{x} d x\) = \((x+1) \tan ^{-1} \sqrt{x}-\sqrt{x}+C .\)

Example 23 Evaluate \(\int \frac{\tan ^{-1} x}{(1+x)^2} d x\)

Solution

Given

\(\int \frac{\tan ^{-1} x}{(1+x)^2} d x\)

Integrating by parts, taking tan^-1x as the first function and \(\frac{1}{(1+x)^2}\) as the second function, we get

I = \(\tan ^{-1} x \cdot \frac{(-1)}{(1+x)}-\int \frac{1}{\left(1+x^2\right)} \cdot \frac{(-1)}{(1+x)} d x\)

= \(\frac{-\tan ^{-1} x}{(1+x)}+\int \frac{d x}{(1+x)\left(1+x^2\right)}=\frac{-\tan ^{-1} x}{(1+x)}+\frac{1}{2} \cdot \int\left\{\frac{1}{(1+x)}+\frac{(1-x)}{\left(1+x^2\right)}\right\}\)

[by partial fractions]

= \(\frac{-\tan ^{-1} x}{(1+x)}+\frac{1}{2} \log |1+x|+\frac{1}{2} \tan ^{-1} x-\frac{1}{4} \log \left(1+x^2\right)+C\).

\(\int \frac{\tan ^{-1} x}{(1+x)^2} d x\) = \(\frac{-\tan ^{-1} x}{(1+x)}+\frac{1}{2} \log |1+x|+\frac{1}{2} \tan ^{-1} x-\frac{1}{4} \log \left(1+x^2\right)+C\).

Integrals of the form \(\int e^x\left[f(x)+f^{\prime}(x)\right] d x\)

Theorem 1 \(\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x \cdot f(x)+C\)

Proof

\(\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=\int e^x \cdot f(x) d x+\int e^x \cdot f^{\prime}(x) d x\)

= \(f(x) \cdot \int e^x d x-\int\left\{f^{\prime}(x) \cdot \int e^x d x\right\} d x+\int e^x f^{\prime}(x) d x+C\)

[evaluating the first integral by parts]

= \(e^x f(x)-\int e^x f^{\prime}(x) d x+\int e^x f^{\prime}(x) d x+C\)

= \(e^x f(x)+C\)

∴ \(\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)+C\)

Integration Techniques 

Example 24 Evaluate:

(1) \(\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) d x\)

(2) \(\int e^x\left(\frac{1}{x^2}-\frac{2}{x^3}\right) d x\)

(3) \(\int e^x\left\{\sin ^{-1} x+\frac{1}{\sqrt{1-x^2}}\right\} d x\)

(4) \(\int e^x(\tan x+\log \sec x) d x\)

Solution We have

(1) I = \(\int e^x\left\{\frac{1}{x}+\left(-\frac{1}{x^2}\right)\right\} d x\)

= \(\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x\), where f(x) = \(\frac{1}{x}\) and f'(x) = \(\frac{-1}{x^2}\)

= \(e^x \cdot f(x)+C=e^x \cdot \frac{1}{x}+C=\frac{e^x}{x}+C\)

(2) I = \(\int e^x\left\{\frac{1}{x^2}+\left(\frac{-2}{x^3}\right)\right\} d x\)

= \(\int e^x\left[f(x)+f^{\prime}(x)\right] d x\), where f(x) = \(\frac{-1}{x^2}\) and f'(x) = \(\frac{-2}{x^3}\)

= \(e^x \cdot f(x)+C=e^x \cdot \frac{1}{x^2}+C=\frac{e^x}{x^2}+C .\)

(3) I = \(\int e^x\left|\sin ^{-1} x+\frac{1}{\sqrt{1-x^2}}\right| d x\)

= \(\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x\), where f(x) = sin-x and f'(x) = \(\frac{1}{\sqrt{1-x^2}}\)

= \(e^x \cdot f(x)+\mathrm{C}=e^x \cdot \sin ^{-1} x+\mathrm{C}=e^x \sin ^{-1} x+\mathrm{C}\)

(4) I = \(\int e^x(\tan x+\log \sec x) d x\)

= \(\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x\), where f(x) = log(sec x)

and f'(x) = \(\frac{1}{\sec x} \cdot \sec x \tan x=\tan x\)

= \(e^x f(x)+C=e^x \log (\sec x)+C\)

Example 25 Evaluate \(\int \frac{x e^x}{(1+x)^2} d x\)

Solution

Given

\(\int \frac{x e^x}{(1+x)^2} d x\)

We have

I = \(\int e^x \cdot\left\{\frac{x}{(1+x)^2}\right\} d x=\int e^x \cdot\left\{\frac{(1+x)-1}{(1+x)^2}\right\} d x\)

= \(\int e^x \cdot\left\{\frac{(1+x)}{(1+x)^2}-\frac{1}{(1+x)^2}\right\} d x=\int e^x \cdot\left\{\frac{1}{(1+x)}-\frac{1}{(1+x)^2}\right\} d x\)

= \(\int e^x \cdot\left\{f(x)+f^{\prime}(x)\right\} d x\), where f(x) = \(\frac{1}{1+x}\) and f'(x) = \(\frac{-1}{(1+x)^2}\)

= \(e^x \cdot f(x)+\mathrm{C}=e^x \cdot \frac{1}{(1+x)}+\mathrm{C}=\frac{e^x}{(1+x)}+\mathrm{C} .\)

\(\int \frac{x e^x}{(1+x)^2} d x\) = \(e^x \cdot f(x)+\mathrm{C}=e^x \cdot \frac{1}{(1+x)}+\mathrm{C}=\frac{e^x}{(1+x)}+\mathrm{C} .\)

Example 26 Evaluate \(\int e^x\left(\frac{1-\sin x}{1-\cos x}\right) d x\)

Solution

Given 

\(\int e^x\left(\frac{1-\sin x}{1-\cos x}\right) d x\)

We have

I = \(\int e^x \cdot\left(\frac{1-\sin x}{1-\cos x}\right) d x=\int e^x \cdot\left\{\frac{1}{(1-\cos x)}-\frac{\sin x}{(1-\cos x)}\right\} d x\)

= \(\int e^x \cdot\left\{\frac{1}{2 \sin ^2(x / 2)}-\frac{2 \sin (x / 2) \cos (x / 2)}{2 \sin ^2(x / 2)}\right\} d x\)

= \(\int e^x \cdot\left\{\frac{1}{2} {cosec}^2 \frac{x}{2}-\cot \frac{x}{2}\right\} d x\)

= \(\int e^x \cdot\left\{-\cot \frac{x}{2}+\frac{1}{2} {cosec}^2 \frac{x}{2}\right\} d x\)

= \(\int e^x \cdot\left\{f(x)+f^{\prime}(x)\right\} d x\),

where f(x) = -cot \(\frac{x}{2}\) and f'(x) = \(\frac{1}{2}\) cosec2 \(\frac{x}{2}\)

= \(e^x \cdot f(x)+C=e^x\left(-\cot \frac{x}{2}\right)+C==-e^x \cot \frac{x}{2}+C .\)

\(\int e^x\left(\frac{1-\sin x}{1-\cos x}\right) d x\) = \(e^x \cdot f(x)+C=e^x\left(-\cot \frac{x}{2}\right)+C==-e^x \cot \frac{x}{2}+C .\)

Example 27 Evaluate \(\int e^x \cdot\left\{\frac{2+\sin 2 x}{1+\cos 2 x}\right\} d x\)

Solution

Given:

\(\int e^x \cdot\left\{\frac{2+\sin 2 x}{1+\cos 2 x}\right\} d x\)

We have

I = \(\int e^x \cdot\left\{\frac{2+\sin 2 x}{1+\cos 2 x}\right\} d x\)

= \(\int e^x \cdot\left\{\frac{2}{(1+\cos 2 x)}+\frac{\sin 2 x}{(1+\cos 2 x)}\right\} d x\)

= \(\int e^x \cdot\left\{\frac{2}{2 \cos ^2 x}+\frac{2 \sin x \cos x}{2 \cos ^2 x}\right\} d x=\int e^x \cdot\left\{\sec ^2 x+\tan x\right\} d x\)

= \(\int e^x \cdot\left\{\tan x+\sec ^2 x \mid d x\right.\)

= \(\int e^x \cdot\left\{f(x)+f^{\prime}(x)\right\} d x\), where f(x) = tan x and f'(x) = sec²x

= e^x . f(x) + C = e^x tan x + C.

Example 28 Evaluate \(\int \frac{\left(x^2+1\right) e^x}{(x+1)^2} d x\)

Solution

Given

\(\int \frac{\left(x^2+1\right) e^x}{(x+1)^2} d x\)

We have

I = \(\int e^x \cdot \frac{\left(x^2+1\right)}{(x+1)^2} d x=\int e^x \cdot\left\{\frac{(x+1)^2-2 x}{(x+1)^2}\right\} d x\)

= \(\int e^x \cdot\left\{1-\frac{2 x}{(x+1)^2}\right\} d x=\int e^x d x-2 \int e^x \cdot \frac{x}{(x+1)^2} d x\)

= \(e^x-2 \cdot \int e^x \cdot \frac{\{(x+1)-1\}}{(x+1)^2} d x=e^x-2 \cdot \int e^x \cdot\left\{\frac{1}{(x+1)}-\frac{1}{(x+1)^2}\right\} d x\)

= \(e^x-2 \cdot \int e^x \cdot\left|f(x)+f^{\prime}(x)\right| d x\), where f(x) = \(\frac{1}{(x+1)}\) and f'(x) = \(\frac{-1}{(x+1)^2}\)

= \(e^x-2 e^x \cdot f(x)+C=e^x-2 e^x \cdot \frac{1}{(x+1)}+C\)

= \(e^x \cdot\left\{1-\frac{2}{(x+1)}\right\}+C=e^x\left(\frac{x-1}{x+1}\right)+C\)

\(\int \frac{\left(x^2+1\right) e^x}{(x+1)^2} d x\) = \(e^x \cdot\left\{1-\frac{2}{(x+1)}\right\}+C=e^x\left(\frac{x-1}{x+1}\right)+C\)

Example 29 Evaluate \(\int e^{2 x}\left(\frac{\sin 4 x-2}{1-\cos 4 x}\right) d x\)

Solution

Given

\(\int e^{2 x}\left(\frac{\sin 4 x-2}{1-\cos 4 x}\right) d x\)

Putting 2x = t and dx= \(\frac{1}{2} dt\), we get

I = \(\frac{1}{2} \int e^t\left(\frac{\sin 2 t-2}{1-\cos 2 t}\right) d t=\frac{1}{2} \int e^t\left(\frac{2 \sin t \cos t-2}{2 \sin ^2 t}\right) d t\)

= \(\frac{1}{2} \int e^t\left\{\frac{\sin t \cos t-1}{\sin ^2 t}\right\} d t=\frac{1}{2} \int e^{\prime}\left(\cot t-{cosec}^2 t\right) d t\)

= \(\frac{1}{2} \int e^t\left\{f(t)+f^{\prime}(t)\right\} d t\), where f(t) = cot t

= \(\frac{1}{2} e^t \cdot f(t)+\mathrm{C}=\frac{1}{2} e^t \cot t+\mathrm{C}\)

= \(\frac{1}{2} e^{2 x} \cot 2 x+C \text {. }\)

\(\int e^{2 x}\left(\frac{\sin 4 x-2}{1-\cos 4 x}\right) d x\) = \(\frac{1}{2} e^{2 x} \cot 2 x+C \text {. }\)

Integrals of the form \(\int e^{k x} \cdot\left\{k \cdot f(x)+f^{\prime}(x)\right\} d x\)

Theorem 2 \(\int e^{k x}\left\{k \cdot f(x)+f^{\prime}(x)\right\} d x=e^{k x} \cdot f(x)+C .\)

Proof

\(\int e^{k x}\left\{k \cdot f(x)+f^{\prime}(x)\right\} d x\)

= \(k \cdot \int e^{k x} f(x) d x+\int e^{k x} f^{\prime}(x) d x\)

= \(k \cdot\left[f(x) \cdot \frac{e^{k x}}{k}-\int f^{\prime}(x) \cdot \frac{e^{k x}}{k} d x\right]+\int e^{k x} f^{\prime}(x) d x+C\)

[evaluating the first integral by parts]

= \(e^{k x} \cdot f(x)-\int e^{k x} \cdot f^{\prime}(x) d x+\int e^{k x} f^{\prime}(x) d x+C=e^{k x} \cdot f(x)+C\)

∴ \(\int e^{k x} \cdot\left\{k \cdot f(x)+f^{\prime}(x)\right\} d x=e^{k x} \cdot f(x)+C .\)

Integration Techniques 

Example 30 Evaluate \(\int e^{2 x} \cdot(-\sin x+2 \cos x) d x\)

Solution

Given

\(\int e^{2 x} \cdot(-\sin x+2 \cos x) d x\)

We have

I = \(\int e^{2 x} \cdot\{2 \cos x-\sin x\} d x=2 \int e^{2 x} \cos x d x-\int e^{2 x} \sin x d x\)

= \(2 \cdot\left[\cos x \cdot \frac{e^{2 x}}{2}-\int(-\sin x) \cdot \frac{e^{2 x}}{2} d x\right]-\int e^{2 x} \sin x d x\)

[integrating e^2x cos x by parts]

= \(e^{2 x} \cos x+\int e^{2 x} \sin x d x-\int e^{2 x} \sin x d x+\mathrm{C}\)

= e^2x cos x + C.

\(\int e^{2 x} \cdot(-\sin x+2 \cos x) d x\) = e^2x cos x + C.

Integrals of the form e^ax cos (bx+c) and e^ax sin (bx+c)

Example 31 Evaluate \(\int e^{a x} \cos (b x+c) d x\)

Solution

Given

\(\int e^{a x} \cos (b x+c) d x\)

Integrating by parts, taking e^ax as the second function, we get

\(\int e^{a x} \cos (b x+c) d x=\cos (b x+c) \cdot \frac{e^{a x}}{a}-\int\left\{-b \sin (b x+c) \cdot \frac{e^{a x}}{a}\right\} d x\)

= \(\frac{e^{a x}}{a} \cos (b x+c)+\frac{b}{a} \int e^{a x} \sin (b x+c) d x\)

= \(\frac{e^{a x}}{a} \cdot \cos (b x+c)+\frac{b}{a}\left[\sin (b x+c) \cdot \frac{e^{\pi x}}{a}-\int\left\{b \cos (b x+c) \cdot \frac{e^{a x}}{a}\right]\right] d x+C\)

[integrating e^ax sin (bx+c) by parts]

= \(\frac{e^{a x}}{a} \cdot \cos (b x+c)+\frac{b}{a^2} e^{a x} \sin (b x+c)-\frac{b^2}{a^2} \int e^{a x} \cos (b x+c) d x+C\)

∴ \(\left(1+\frac{b^2}{a^2}\right) \int e^{a x} \cos (b x+c) d x=\frac{e^{a x}}{a} \cos (b x+c)+\frac{b}{a^2} e^{a x} \sin (b x+c)+C\)

or \(\int e^{a x} \cos (b x+c) d x=e^{a x}\left[\frac{a \cos (b x+c)+b \sin (b x+c)}{\left(a^2+b^2\right)}\right]+C^{\prime} .\)

Remark

Put a = r cos θ and b = r sin θ so that

r = \(\sqrt{a^2+b^2}\) and θ = \(\tan ^{-1}\left(\frac{b}{a}\right).\)

∴ \(\int e^{a x} \cos (b x+c) d x=\frac{r e^{a x} \cos (b x+c-\theta)}{\left(a^2+b^2\right)}\)

= \(e^{a x} \cdot \frac{\cos \left[b x+c-\tan ^{-1}(b / a)\right]}{\sqrt{a^2+b^2}} .\)

Similarly, \(\int e^{a x} \sin (b x+c) d x=e^{a x} \cdot \frac{\sin \left[b x+c-\tan ^{-1}(b / a)\right]}{\sqrt{a^2+b^2}}\).

WBCHSE Class 12 Maths Solutions Indefinite Integral

Integration It is the inverse process of differentiation.

If the derivative of F(x) is f(x) then we say that the antiderivative or integral of f(x) is F(x) and we write,

\(\int f(x) d x=F(x) .\)

Thus, \(\frac{d}{d x}[F(x)]=f(x) \Rightarrow \int f(x) d x=F(x) .\)

Example Since \(\frac{d}{d x}(\sin x)=\cos x\), we have \(\int \cos x d x=\sin x.\)

Moreover, if C is any constant then \(\frac{d}{d x}(\sin x+C)=\cos x.\)

So, in general, \(\int \cos x d x=(\sin x+C) .\)

Different values of C will give different integrals.

Real-Life Applications of Indefinite Integrals

Indefinite Integral Examples And Solutions

Thus, a given function may have an indefinite number of integrals. Because of this property, we call these integrals indefinite integrals.

Thus, \(\frac{d}{d x}[F(x)]=f(x) \Rightarrow \int f(x) d x=F(x)+C\), where C is a constant, called the constant of integration. Any function to be integrated is known as an integrand.

The following two results are a direct consequence of the definition of an integral.

Result 1 \(\int x^n d x=\frac{x^{(n+1)}}{(n+1)}+C\), when n ≠ -1.

Proof

We have, \(\frac{d}{d x}\left(\frac{x^{n+1}}{n+1}\right)=\frac{(n+1) x^n}{(n+1)}=x^n\)

∴ \(\int x^n d x=\frac{x^{(n+1)}}{(n+1)}+\text { C. }\)

Thus, we have

Read and Learn More  Class 12 Math Solutions

(1) \(\int x^6 d x=\frac{x^{(6+1)}}{(6+1)}+C=\frac{x^7}{7}+C .\)

(2) \(\int x^{2 / 3} d x=\frac{x^{\left(\frac{2}{3}+1\right)}}{\left(\frac{2}{3}+1\right)}+C=\frac{3}{5} x^{5 / 3}+C\)

(3) \(\int x^{-3 / 4} d x=\frac{x^{\left(-\frac{3}{4}+1\right)}}{\left(-\frac{3}{4}+1\right)}=4 x^{1 / 4}+C\)

Result 2 \(\int \frac{1}{x} d x=\log |x|+C\), where x ≠ 0.

Proof

Either x > 0 or x < 0.

Case 1 When x > 0

In this case, | x | = x.

∴ \(\frac{d}{d x}[\log |x|]=\frac{d}{d x}(\log x)=\frac{1}{x}\)

Indefinite Integral Examples And Solutions

So, we have, \(\int \frac{1}{x} d x=\log |x|+C \text {. }\)

Case 2 When x < 0

In this case | x | = -x.

∴ \(\frac{d}{d x}[\log |x|]=\frac{d}{d x}[\log (-x)]=\frac{1}{(-x)} \cdot(-1)=\frac{1}{x}\)

So, we have \(\int \frac{1}{x} d x=\log |x|+C .\)

Thus, from both the cases, we have \(\int \frac{1}{x} d x=\log |x|+C\)

WBCHSE Class 12 Maths Solutions

Formulae

On the basis of differentiation and the definition of integration, we have the following results.

1. \(\frac{d}{d x}\left(\frac{x^{n+1}}{n+1}\right)=x^n, n \neq-1 \Rightarrow \int x^n d x=\frac{x^{n+1}}{(n+1)}+C\)

2. \(\frac{d}{d x}(\log |x|)=\frac{1}{x} \Rightarrow \int \frac{1}{x} d x=\log |x|+C\)

3. \(\frac{d}{d x}\left(e^x\right)=e^x \Rightarrow \int e^x d x=e^x+\mathrm{C}\)

4. \(\frac{d}{d x}\left(\frac{a^x}{\log a}\right)=a^x \Rightarrow \int a^x d x=\frac{a^x}{\log a}+C\)

5. \(\frac{d}{d x}(\sin x)=\cos x \Rightarrow \int \cos x d x=\sin x+C\)

Indefinite Integral Examples And Solutions

6. \(\frac{d}{d x}(-\cos x)=\sin x \Rightarrow \int \sin x d x=-\cos x+C\)

7. \(\frac{d}{d x}(\tan x)=\sec ^2 x \Rightarrow \int \sec ^2 x d x=\tan x+C\)

8. \(\frac{d}{d x}(-\cot x)={cosec}^2 x \Rightarrow \int {cosec}^2 x d x=-\cot x+C\)

9. \(\frac{d}{d x}(\sec x)=\sec x \tan x \Rightarrow \int \sec x \tan x d x=\sec x+C\)

10. \(\frac{d}{d x}(-{cosec} x)={cosec} x \cot x \Rightarrow \int {cosec} x \cot x d x=-{cosec} x+C\)

11. \(\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^2}} \Rightarrow \int \frac{1}{\sqrt{1-x^2}} d x=\sin ^{-1} x+C\)

12. \(\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{\left(1+x^2\right)} \Rightarrow \int \frac{1}{\left(1+x^2\right)} d x=\tan ^{-1} x+C\)

13. \(\frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{x \sqrt{x^2-1}} \Rightarrow \int \frac{1}{x \sqrt{x^2-1}} d x=\sec ^{-1} x+C\)

With the help of the above formulae, it is easy to evaluate the following integrals.

WBBSE Class 12 Indefinite Integrals Solutions

Example 1 Evaluate:

(1) \(\int x^9 d x\)

(2) \(\int \sqrt[3]{x} d x\)

(3) \(\int d x\)

(4) \(\int \frac{1}{x^2} d x\)

(5) \(\int \frac{1}{x^{1 / 3}} d x\)

(6) \(\int 5^x d x\)

Solution

Using the standard formulae, we have

(1) \(\int x^9 d x=\frac{x^{(9+1)}}{(9+1)}+C=\frac{x^{10}}{10}+C\)

(2) \(\int \sqrt[3]{x} d x=\int x^{1 / 3} d x=\frac{x^{\left(\frac{1}{3}+1\right)}}{\left(\frac{1}{3}+1\right)}+\mathrm{C}=\frac{3}{4} x^{4 / 3}+\mathrm{C} .\)

Indefinite Integration All Formulas

(3) \(\int d x=\int x^0 d x=\frac{x^{(0+1)}}{(0+1)}+C=x+C\)

(4) \(\int \frac{1}{x^2} d x=\int x^{-2} d x=\frac{x^{(-2+1)}}{(-2+1)}+C=-\frac{1}{x}+C .\)

(5) \(\int \frac{1}{x^{1 / 3}} d x=\int x^{-1 / 3} d x=\frac{x^{\left(-\frac{1}{3}+1\right)}}{\left(-\frac{1}{3}+1\right)}+C=\frac{3}{2} x^{2 / 3}+C\)

(6) \(\int 5^x d x=\frac{5^x}{\log 5}+C\)

Examples of Integration by Substitution Method

Some Standard Results on Integration

Theorem 1 \(\frac{d}{d x}\left\{\int f(x) d x\right\}=f(x) .\)

Proof

Let \(\int f(x) d x=F(x)\) …(1)

Then, \(\frac{d}{d x}\{F(x)\}=f(x)\) [by def. of integral].

∴ \(\frac{d}{d x}\left\{\int f(x) d x\right\}=f(x)\) [using (1)]

Theorem 2 \(\int k \cdot f(x) d x=k \cdot \int f(x) d x\), where k is a constant.

Proof

Let \(\int f(x) d x=F(x)\) …(1)

Then, \(\frac{d}{d x}\{F(x)\}=f(x)\) …(2)

∴ \(\frac{d}{d x}\{k \cdot F(x)\}=k \cdot \frac{d}{d x}\{F(x)\}=k \cdot f(x)\) [using(2)]

So, by the definition of an integral, we have

\(\int\{k \cdot f(x)\} d x=k \cdot F(x)=k \cdot \int f(x) d x\) [using (1)].

Example 2 Evaluate:

(1) \(\int 3 x^2 d x\)

(2) \(\int 2^{(x+3)} d x\)

Solution

(1) \(\int 3 x^2 d x=3 \int x^2 d x=3 \cdot \frac{x^3}{3}+C=x^3+C .\)

Indefinite Integration All Formulas

(2) \(\int 2^{(x+3)} d x=\int 2^x \cdot 2^3 d x=8 \int 2^x d x=8 \cdot \frac{2^x}{\log 2}+C=\frac{2^{(x+3)}}{\log 2}+C .\)

Theorem 3

(1) \(\int\left\{f_1(x)+f_2(x)\right\} d x=\int f_1(x) d x+\int f_2(x) d x\)

(2) \(\int\left\{f_1(x)-f_2(x)\right\} d x=\int f_1(x) d x-\int f_2(x) d x\)

Proof

(1) Let \(\int f_1(x) d x=F_1(x) \text { and } \int f_2(x) d x=F_2(x)\) …(1)

Then, \(\frac{d}{d x}\left\{F_1(x)\right\}=f_1(x) \text { and } \frac{d}{d x}\left\{F_2(x)\right\}=f_2(x)\) …(2)

Now, \(\frac{d}{d x}\left\{F_1(x)+F_2(x)\right\}=\frac{d}{d x}\left\{F_1(x)\right\}+\frac{d}{d x}\left\{F_2(x)\right\}\)

= f₁(x) + f₂(x) [using(2)].

∴ \(\int\left\{f_1(x)+f_2(x)\right] d x=F_1(x)+F_2(x)\)

= \(\int f_1(x) d x+\int f_2(x) d x\) [using (1)].

Similarly, (2) can be proved.

Indefinite Integration All Formulas

Remark In general, we have

\(\int\left\{k_1 \cdot f_1(x) \pm k_2 \cdot f_2(x) \pm \ldots \pm k_n \cdot f_n(x)\right\} d x = k_1 \cdot \int f_1(x) d x \pm k_2 \cdot\) \(\int f_2(x) d x \pm \ldots \pm k_n \cdot \int f_n(x) d x\)

Solved Examples

Example 1 Evaluate:

(1) \(\int\left(5 x^3+2 x^{-5}-7 x+\frac{1}{\sqrt{x}}+\frac{5}{x}\right) d x\)

(2) \(\int\left(3 \sin x-4 \cos x+5 \sec ^2 x-2 {cosec}^2 x\right) d x\)

(3) \(\int(1-x)(2+3 x)(5-4 x) d x\)

(4) \(\int\left(\frac{3 x^4-5 x^3+4 x^2-x+2}{x^3}\right) d x\)

(5) \(\int\left(x^2+\frac{1}{x^2}\right)^3 d x\)

Solution

(1) \(\int\left(5 x^3+2 x^{-5}-7 x+\frac{1}{\sqrt{x}}+\frac{5}{x}\right) d x\)

= \(5 \int x^3 d x+2 \int x^{-5} d x-7 \int x d x+\int x^{-1 / 2} d x+5 \int \frac{1}{x} d x\)

= \(5 \cdot \frac{x^4}{4}+2 \cdot \frac{x^{-4}}{(-4)}-7 \cdot \frac{x^2}{2}+\frac{x^{1 / 2}}{(1 / 2)}+5 \log |x|+C\)

= \(\frac{5 x^4}{4}-\frac{1}{2 x^4}-\frac{7 x^2}{2}+2 \sqrt{x}+5 \log |x|+C\)

(2) \(\int\left(3 \sin x-4 \cos x+5 \sec ^2 x-2 {cosec}^2 x\right) d x\)

= \(3 \int \sin x d x-4 \int \cos x d x+5 \int \sec ^2 x d x-2 \int {cosec}^2 x d x\)

= 3(-cos x) – 4 sin x + 5 tan x – 2 (-cot x) + C

= (-3 cos x – 4 sin x + 5 tan x + 2 cot x + C).

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Properties of Indefinite Integrals Explained

Indefinite Integration Properties

(3) \(\int(1-x)(2+3 x)(5-4 x) d x\)

= \(\int\left(10-3 x-19 x^2+12 x^3\right) d x\)

= \(10 \int d x-3 \int x d x-19 \int x^2 d x+12 \int x^3 d x\)

= \(10 x-3 \cdot \frac{x^2}{2}-19 \cdot \frac{x^3}{3}+12 \cdot \frac{x^4}{4}+C\)

= \(10 x-\frac{3 x^2}{2}-\frac{19 x^3}{3}+3 x^4+C\)

(4) \(\int\left(\frac{3 x^4-5 x^3+4 x^2-x+2}{x^3}\right) d x\)

= \(\int\left(3 x-5+\frac{4}{x}-\frac{1}{x^2}+\frac{2}{x^3}\right) d x\) [dividing each term by x³]

= \(3 \int x d x-5 \int d x+4 \int \frac{1}{x} d x-\int x^{-2} d x+2 \int x^{-3} d x\)

= \(3 \cdot \frac{x^2}{2}-5 x+4 \log |x|-\left(-\frac{1}{x}\right)+2\left(\frac{x^{-2}}{-2}\right)+C\)

= \(\frac{3 x^2}{2}-5 x+4 \log |x|+\frac{1}{x}-\frac{1}{x^2}+C\)

(5) \(\int\left(x^2+\frac{1}{x^2}\right)^3 d x\)

= \(\int\left(x^6+\frac{1}{x^6}+3 x^2+\frac{3}{x^2}\right) d x\)

= \(\int x^6 d x+\int x^{-6} d x+3 \int x^2 d x+3 \int \frac{1}{x^2} d x\)

= \(\frac{x^7}{7}+\frac{x^{-5}}{(-5)}+3 \cdot \frac{x^3}{3}+3 \cdot\left(-\frac{1}{x}\right)+C\)

= \(\frac{x^7}{7}-\frac{1}{5 x^5}+x^3-\frac{3}{x}+\mathrm{C} .\)

Example 2 Evaluate:

(1) \(\int \frac{\left(x^3+4 x^2-3 x-2\right)}{(x+2)} d x\)

(2) \(\int\left(\frac{x^4+1}{x^2+1}\right) d x\)

Indefinite Integration Properties

Solution

(1) On dividing (x³ + 4x² – 3x – 2) by (x + 2), we get

\(\int \frac{\left(x^3+4 x^2-3 x-2\right)}{(x+2)} d x=\int\left\{x^2+2 x-7+\frac{12}{x+2}\right\} d x\)

= \(\int x^2 d x+2 \int x d x-7 \int d x+12 \int \frac{1}{x+2} d x\)

= \(\frac{x^3}{3}+2 \cdot \frac{x^2}{2}-7 x+12 \log |x+2|+C\)

= \(\frac{x^3}{3}+x^2-7 x+12 \log |x+2|+C\)

(2) On dividing (x⁴ + 1) by (x² + 1), we get

\(\int\left(\frac{x^4+1}{x^2+1}\right) d x=\int\left[x^2-1+\frac{2}{\left(x^2+1\right)}\right] d x\)

= \(\int x^2 d x-\int d x+2 \int \frac{1}{x^2+1} d x=\frac{x^3}{3}-x+2 \tan ^{-1} x+\mathrm{C} .\)

Example 3 Evaluate:

(1) \(\int \tan ^2 x d x\)

(2) \(\int \cot ^2 x d x\)

(3) \(\int \sin ^2 \frac{x}{2} d x\)

Solution

(1) \(\int \tan ^2 x d x=\int\left(\sec ^2 x-1\right) d x\)

= \(\int \sec ^2 x d x-\int d x=\tan x-x+C\)

(2) \(\int \cot ^2 x d x=\int\left({cosec}^2 x-1\right) d x\)

= \(\int {cosec}^2 x d x-\int d x=-\cot x-x+C\)

(3) We know that \(2 \sin ^2 \frac{x}{2}=(1-\cos x)\)

∴ \(\int \sin ^2 \frac{x}{2} d x=\frac{1}{2} \int(1-\cos x) d x\)

= \(\frac{1}{2}\left[\int d x-\int \cos x d x\right]=\frac{1}{2} x-\frac{1}{2} \sin x+C .\)

Integration Techniques for Indefinite Integrals

Example 4 Evaluate \(\int \sqrt{1-\sin 2 x} d x\)

Solution

\(\int \sqrt{1-\sin 2 x} d x=\int\left(\cos ^2 x+\sin ^2 x-2 \sin x \cos x\right)^{1 / 2} d x\)

= \(\int \sqrt{(\cos x-\sin x)^2} d x\)

= \(\int(\cos x-\sin x) d x=\int \cos x d x-\int \sin x d x\)

= sin x – (- cos x) + C = sin x + cos x + C.

\(\int \sqrt{1-\sin 2 x} d x\) = sin x – (- cos x) + C = sin x + cos x + C.

Common Formulas for Indefinite Integrals

Indefinite Integration Properties

Example 5 Evaluate:

(1) \(\int \frac{d x}{1+\sin x}\)

(2) \(\int\left(\frac{\sin x}{1+\sin x}\right) d x\)

Solution

(1) \(\int \frac{d x}{(1+\sin x)}=\int \frac{1}{(1+\sin x)} \times \frac{(1-\sin x)}{(1-\sin x)} d x\)

= \(\int \frac{(1-\sin x)}{\left(1-\sin ^2 x\right)} d x=\int \frac{(1-\sin x)}{\cos ^2 x} d x\)

= \(\int\left(\frac{1}{\cos ^2 x}-\frac{\sin x}{\cos ^2 x}\right) d x=\int\left(\sec ^2 x-\sec x \tan x\right) d x\)

= \(\int \sec ^2 x d x-\int \sec x \tan x d x=\tan x-\sec x+C\)

(2) \(\int\left(\frac{\sin x}{1+\sin x}\right) d x=\int \frac{(1+\sin x)-1}{(1+\sin x)} d x\)

= \(\int\left(1-\frac{1}{1+\sin x}\right) d x=\int d x-\int \frac{1}{(1+\sin x)} d x\)

= \(\int d x-\int \frac{1}{(1+\sin x)} \times \frac{(1-\sin x)}{(1-\sin x)} d x\)

= \(\int d x-\int \frac{(1-\sin x)}{\cos ^2 x} d x=\int d x-\int\left(\frac{1}{\cos ^2 x}-\frac{\sin x}{\cos ^2 x}\right) d x\)

= \(\int d x-\int \sec ^2 x d x+\int \sec x \tan x d x=x-\tan x+\sec x+C \text {. }\)

Example 6 Evaluate \(\int \frac{\sec x}{(\sec x+\tan x)} d x\)

Solution

Given

\(\int \frac{\sec x}{(\sec x+\tan x)} d x\) \(\int \frac{\sec x}{(\sec x+\tan x)} d x=\int \frac{\sec x}{(\sec x+\tan x)} \times \frac{(\sec x-\tan x)}{(\sec x-\tan x)} d x\)

= \(\int \frac{\left(\sec ^2 x-\sec x \tan x\right)}{\left(\sec ^2 x-\tan ^2 x\right)} d x\)

= \(\int\left(\sec ^2 x-\sec x \tan x\right) d x\)

Indefinite Integration Properties

= \(\int \sec ^2 x d x-\int \sec x \tan x d x=\tan x-\sec x+C\)

\(\int \frac{\sec x}{(\sec x+\tan x)} d x\) = \(\int \sec ^2 x d x-\int \sec x \tan x d x=\tan x-\sec x+C\)

Example 7 Evaluate:

(1) \(\int\left(\frac{4-5 \cos x}{\sin ^2 x}\right) d x\)

(2) \(\int\left(\frac{1-\cos 2 x}{1+\cos 2 x}\right) d x\)

(3) \(\int \frac{1}{\sin ^2 x \cos ^2 x} d x\)

(4) \(\int \frac{\cos 2 x}{\cos ^2 x \sin ^2 x} d x\)

Solution

(1) \(\int\left(\frac{4-5 \cos x}{\sin ^2 x}\right) d x=\int\left(\frac{4}{\sin ^2 x}-\frac{5 \cos x}{\sin ^2 x}\right) d x\)

= \(\int\left(4 {cosec}^2 x-5 {cosec} x \cot x\right) d x\)

= \(4 \int {cosec}^2 x d x-5 \int {cosec} x \cot x d x\)

= 4(-cot x) – 5(- cosec x) + C

= -4 cot x + 5 cosec x + C.

(2) \(\int\left(\frac{1-\cos 2 x}{1+\cos 2 x}\right) d x=\int \frac{2 \sin ^2 x}{2 \cos ^2 x} d x=\int \tan ^2 x d x\)

= \(\int\left(\sec ^2 x-1\right) d x=\int \sec ^2 x d x-\int d x\)

= tan x – x + C.

(3) \(\int \frac{1}{\sin ^2 x \cos ^2 x} d x=\int\left(\frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x \cos ^2 x}\right) d x\)

= \(\int\left(\frac{1}{\cos ^2 x}+\frac{1}{\sin ^2 x}\right) d x\)

= \(\int \sec ^2 x d x+\int {cosec}^2 x d x=\tan x-\cot x+C\)

(4) \(\int \frac{\cos 2 x}{\cos ^2 x \sin ^2 x} d x=\int\left(\frac{\cos ^2 x-\sin ^2 x}{\cos ^2 x \sin ^2 x}\right) d x\)

= \(\int\left(\frac{1}{\sin ^2 x}-\frac{1}{\cos ^2 x}\right) d x\)

= \(\int {cosec}^2 x d x-\int \sec ^2 x d x=-\cot x-\tan x+C\)

Step-by-Step Solutions to Indefinite Integral Problems

Example 8 Evaluate \(\int\left(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\right) d x\)

Solution

Given:

\(\int\left(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\right) d x\) \(\int\left(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\right) d x=\int \frac{\left(2 \cos ^2 x-1\right)-\left(2 \cos ^2 \alpha-1\right)}{(\cos x-\cos \alpha)} d x\)

= \(2 \int \frac{\left(\cos ^2 x-\cos ^2 \alpha\right)}{(\cos x-\cos \alpha)} d x=2 \int(\cos x+\cos \alpha) d x\)

= \(2 \int \cos x d x+2 \cos \alpha \cdot \int d x=2 \sin x+2 x \cos \alpha+C \text {. }\)

\(\int\left(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\right) d x\) = \(2 \int \cos x d x+2 \cos \alpha \cdot \int d x=2 \sin x+2 x \cos \alpha+C \text {. }\)

Example 9 Evaluate \(\int \tan ^{-1}\left\{\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right\} d x\)

Solution

Given

\(\int \tan ^{-1}\left\{\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right\} d x\) \(\int \tan ^{-1}\left\{\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right\} d x=\int \tan ^{-1}\left\{\sqrt{\frac{2 \sin ^2 x}{2 \cos ^2 x}}\right\} d x\)

= \(\int \tan ^{-1}(\tan x) d x=\int x d x=\frac{x^2}{2}+C\)

\(\int \tan ^{-1}\left\{\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right\} d x\) = \(\int \tan ^{-1}(\tan x) d x=\int x d x=\frac{x^2}{2}+C\)

Example 10 Evaluate \(\int \sin ^{-1}(\cos x) d x\)

Solution

Given

\(\int \sin ^{-1}(\cos x) d x\) \(\int \sin ^{-1}(\cos x) d x=\int \sin ^{-1}\left\{\sin \left(\frac{\pi}{2}-x\right)\right\} d x\)

= \(\int\left(\frac{\pi}{2}-x\right) d x=\frac{\pi}{2} \cdot \int d x-\int x d x=\frac{\pi x}{2}-\frac{x^2}{2}+C\)

\(\int \sin ^{-1}(\cos x) d x\) = \(\int\left(\frac{\pi}{2}-x\right) d x=\frac{\pi}{2} \cdot \int d x-\int x d x=\frac{\pi x}{2}-\frac{x^2}{2}+C\)

Example 11 Evaluate \(\int \tan ^{-1}(\sec x+\tan x) d x\)

Solution

Given

\(\int \tan ^{-1}(\sec x+\tan x) d x\) \(\int \tan ^{-1}(\sec x+\tan x) d x=\int \tan ^{-1}\left(\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right) d x\)

= \(\int \tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right) d x=\int \tan ^{-1}\left\{\frac{1-\cos \left(\frac{\pi}{2}+x\right)}{\sin \left(\frac{\pi}{2}+x\right)}\right\} d x\)

= \(\int \tan ^{-1}\left\{\frac{2 \sin ^2\left(\frac{\pi}{4}+\frac{x}{2}\right)}{2 \sin \left(\frac{\pi}{4}+\frac{x}{2}\right) \cos \left(\frac{\pi}{4}+\frac{x}{2}\right)}\right\} d x\)

= \(\int \tan ^{-1}\left\{\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right\} d x=\int\left(\frac{\pi}{4}+\frac{x}{2}\right) d x\)

= \(\frac{\pi}{4} \cdot \int d x+\frac{1}{2} \int x d x=\frac{\pi x}{4}+\frac{1}{4} x^2+C .\)

\(\int \tan ^{-1}(\sec x+\tan x) d x\) = \(\frac{\pi}{4} \cdot \int d x+\frac{1}{2} \int x d x=\frac{\pi x}{4}+\frac{1}{4} x^2+C .\)

Example 12 Evaluate \(\int\left(\frac{1+\sin x}{1-\sin x}\right) d x\)

Solution

Given

\(\int\left(\frac{1+\sin x}{1-\sin x}\right) d x\)

I = \(\int \frac{(1+\sin x)}{(1-\sin x)} \times \frac{(1+\sin x)}{(1+\sin x)} d x\)

= \(\int \frac{(1+\sin x)^2}{\left(1-\sin ^2 x\right)} d x=\int \frac{\left(1+\sin ^2 x+2 \sin x\right)}{\cos ^2 x} d x\)

= \(\int\left(\frac{1}{\cos ^2 x}+\frac{\sin ^2 x}{\cos ^2 x}+\frac{2 \sin x}{\cos ^2 x}\right) d x=\int\left(\sec ^2 x+\tan ^2 x+2 \sec x \tan x\right) d x\)

= \(\int\left(2 \sec ^2 x-1+2 \sec x \tan x\right) d x\)

= \(2 \int \sec ^2 x d x-\int d x+2 \int \sec x \tan x d x\)

= 2 tan x – x + 2 sec x + C.

\(\int\left(\frac{1+\sin x}{1-\sin x}\right) d x\) = 2 tan x – x + 2 sec x + C.