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

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

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 ^{3} x \cdot \cos x} d x

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

The exponent of denominator is 3+1=4 to solve this type of integral; uses have divided both numerator and denominator by \cos ^{4}x then

\begin{aligned} I &=\int \frac{\frac{1}{\cos ^{4} x}}{\frac{\sin ^{3} x \cdot \cos x}{\cos ^{4} x}} d x \\ &=\int \frac{\left(\sec ^{4} x\right)}{\frac{\sin ^{3} x}{\cos ^{3} x}} d x \end{aligned}
\begin{aligned} &=\int \frac{\sec ^{4} x}{\tan ^{3} x} d x=\int \frac{\sec ^{2} x \cdot \sec ^{2} x}{\tan ^{3} x} d x \\ &=\int \frac{\left(1+\tan ^{2} x\right) \sec ^{2} x}{\tan ^{3} x} d x\quad\quad\quad\quad\quad\quad\quad\quad\left[\because 1+\tan ^{2} x=\sec ^{2} x\right] \\ \end{aligned}    
Substitute \tan x=t \Rightarrow \sec ^{2} x d x=d t then

\begin{aligned} I &=\int \frac{1+t^{2}}{t^{3}} d t \\ &=\int\left(\frac{1}{t^{3}}+\frac{t^{2}}{t^{3}}\right) d t \\ &=\int\left(t^{-3}+\frac{1}{t}\right) d t \end{aligned}
    =\frac{t^{-3+1}}{-3+1}+\log |t|+C \quad\quad\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=x \frac{n+1}{n+1}+c \text { and } \int \frac{1}{x} d x=\log |x|+C\right]
    \begin{aligned} &=\frac{t^{-2}}{-2}+\log |t|+C \\ &=-\frac{1}{2 t^{2}}+\log |t|+C \\ &=-\frac{1}{2 \tan ^{2} x}+\log |\tan x|+C \quad\quad\quad\quad\quad\quad\quad[\because t=\tan x] \end{aligned}

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