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  • Provide Solutio for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.12 Question 13

Provide Solutio for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.12 Question 13

Answers (1)

Answer:\frac{1}{2} \tan ^{2} x+\log |\tan x|+C

Hint: Use substitution method to solve this integral.

Given: \int \frac{1}{\sin x \cdot \cos ^{3} x} d x

Solution: Let I=\int \frac{1}{\sin x \cdot \cos ^{3} x} d x

Re-writing,

\begin{aligned} I &=\int \frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cdot \cos ^{3} x} d x \quad\left[\therefore 1=\sin ^{2} x+\cos ^{2} x\right] \\ &=\int\left(\frac{\sin ^{2} x}{\sin x \cos ^{3} x}+\frac{\cos ^{2} x}{\sin x \cdot \cos ^{3} x}\right) d x \end{aligned}
    \begin{aligned} &=\int\left(\frac{\sin x}{\cos x} \cdot \frac{1}{\cos ^{2} x}+\frac{1}{\sin x \cdot \cos x}\right) d x \\ &=\int\left(\tan x \cdot \sec ^{2} x+\frac{\cos ^{2} x}{\cos ^{2} x} \cdot \frac{1}{\sin x \cdot \cos x}\right) d x \end{aligned}
    =\int\left(\tan x \cdot \sec ^{2} x+\frac{1}{\cos ^{2} x} \cdot \frac{1}{\frac{\sin x \cdot \cos x}{\cos ^{2} x}}\right) d x
    =\int\left(\tan x \cdot \sec ^{2} x+\sec ^{2} x \cdot \frac{1}{\frac{\sin x}{\cos x}}\right) d x
    \begin{aligned} I &=\int\left(\tan x \cdot \sec ^{2} x+\frac{\sec ^{2} x}{\tan x}\right) d x \\ & \Rightarrow I=\int \tan x \cdot \sec ^{2} x d x+\int \frac{\sec ^{2} x}{\tan x} d x \end{aligned}

Substitute \tan x=t \Rightarrow \sec ^{2} x d x=d t then

\begin{aligned} I &=\int t d t+\int_{t}^{1} d t \\ &=\frac{t^{1+1}}{1+1}+\log |t|+C \quad \quad \quad \quad \quad \quad \quad \quad\left[\because \int x^{2} d x=\frac{x^{n+1}}{n+1}+c, \int \frac{1}{x} d x=1\right] \end{aligned}
   \begin{aligned} &=\frac{t^{2}}{2}+\log |t|+C \\ &=\frac{\tan ^{2} x}{2}+\log |\tan x|+C \quad\quad\quad\quad\quad\quad\quad[\because t=\tan x] \end{aligned}

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