Integration – Definition, Types, Methods, Examples

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 xn-1.f(xn), we put xn = t and nxn-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, \(\int \sin ^3 x d x=\int\left(\frac{3}{4} \sin x-\frac{1}{4} \sin 3 x\right) d x\)

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

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

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

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 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

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

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

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.

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

Integration Using Trigonometric Identities

When the integrand consists of trogonometric 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) \(\int \sin ^7 x d x=\int \sin ^6 x \cdot \sin x d x\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 = tan2-1(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

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

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

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

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

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 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

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

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

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

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

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

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]

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

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

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

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.

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 e2x 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]

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

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

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

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

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

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

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

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

= \(\left(\sin ^{-1} x\right) \cdot \frac{x}{\sqrt{1-x^2}}+\log \left|\sqrt{1-x^2}\right|+C\)

[∵ \(\cos t=\sqrt{1-x^2} \text { and } \tan t=\frac{x}{\sqrt{1-x^2}}\)]

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

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

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 = et and \frac{1}{x}dx = dt or dx = etdt.

∴ \(\int \sin (\log x) d x=\int e^t \sin t d t\) ..(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

\(\int \sin (\log x) d x=\int e^t \sin t d t\)

= \(\frac{1}{2}\left(-e^t \cos t+e^t \sin t\right)+C\)

= \(\frac{1}{2}[-x \cos (\log x)+x \sin (\log x)]+C\)

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

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

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]

= \(e^x f(x)-\int e^x f^{\prime}(x) d x+\int e^x f^{\prime}(x) d x+C\)

= \(e^x f(x)+C\)

∴ \(\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)+C\)

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

= ex . f(x) + C = ex 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]

= \(e^{k x} \cdot f(x)-\int e^{k x} \cdot f^{\prime}(x) d x+\int e^{k x} f^{\prime}(x) d x+C=e^{k x} \cdot f(x)+C\)

∴ \(\int e^{k x} \cdot\left\{k \cdot f(x)+f^{\prime}(x)\right\} d x=e^{k x} \cdot f(x)+C .\)

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 e2x cos x by parts]

= \(e^{2 x} \cos x+\int e^{2 x} \sin x d x-\int e^{2 x} \sin x d x+\mathrm{C}\)

= e2x cos x + C.

\(\int e^{2 x} \cdot(-\sin x+2 \cos x) d x\) = e2x cos x + C.

Integrals of the form eax cos (bx+c) and eax 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 eax 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 eax 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, \(\int e^{a x} \sin (b x+c) d x=e^{a x} \cdot \frac{\sin \left[b x+c-\tan ^{-1}(b / a)\right]}{\sqrt{a^2+b^2}}\).

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