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.