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Provide Solutio for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.12 Question 11

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Answer: - -\frac{1}{2} \tan ^{-2} x+3 \log |\tan x|+\frac{3}{2} \tan ^{2} x+\frac{1}{4} \tan ^{4} x+C

Hint: - Use substitution method to solve this integral.

Given: - \int \frac{1}{\sin ^{3} x \cdot \cos ^{5} x} d x

The exponent of denominator is 3+5=8. To solve this type of integral, we have to divide both numerator and denominator by \cos ^{8 }x then

\begin{aligned} &I=\int \frac{\frac{1}{\cos ^{8} x}}{\frac{\sin ^{3} x \cdot \cos ^{5} x}{\cos ^{8} x}} d x=\int \frac{\sec ^{8} x}{\frac{\sin ^{3} x}{\cos ^{3} x}} d x \\ &=\int \frac{\sec ^{8} x}{\tan ^{3} x} d x=\int \frac{\sec ^{6} x \cdot \sec ^{2} x}{\tan ^{3} x} d x=\int \frac{\left(\sec ^{2} x\right)^{3} \sec ^{2} x}{\tan ^{3} x} \end{aligned}
 \begin{aligned} &=\int \frac{\left(1+\tan ^{2} x\right)^{3} \sec ^{2} x}{\tan ^{2} x} d x \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\left[\because 1+\tan ^{2} x=\sec ^{2} x\right] \\ &\Rightarrow I=\int \frac{\left(1+\tan ^{6} x+3 \tan ^{2} x+3 \tan ^{4} x\right) \sec ^{2} x}{\tan ^{3} x} d x \end{aligned}
Substitute \tan x=t \Rightarrow \sec ^{3} x d x=d t then

\begin{aligned} I &=\int \frac{\left(1+t^{6}+3 t^{2}+3 t^{4}\right)}{t^{3}} d t=\int\left(\frac{1}{t^{3}}+\frac{t^{6}}{t^{3}}+\frac{3 t^{2}}{t^{3}}+\frac{3 t^{4}}{t^{3}}\right) d t \\ &=\int\left(t^{-3}+t^{6-3}+\frac{3}{t}+3 t\right) d t=\int\left(t^{-3}+t^{3}+\frac{3}{t}+3 t\right) d t \\ &=\int t^{-3} d t+\int t^{3} d t+3 \int_{t}^{1} \frac{1}{t} d t+3 \int t d t \end{aligned}
    =\frac{t^{-3+1}}{-3+1}+\frac{t^{3+1}}{3+1}+3 \log |t|+\frac{3 t^{1+1}}{1+1}+C \quad\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \& \int \frac{1}{t} d t=\log |t|+C\right]
    \begin{aligned} &=\frac{t^{-2}}{-2}+\frac{t^{4}}{4}+3 \log |t|+\frac{3 t^{2}}{2}+C \\ &=-\frac{1}{2} t^{-2}+3 \log |t|+\frac{3 t^{2}}{2}+\frac{1}{4} t^{4}+C \\ &=-\frac{1}{2 \tan ^{2} x}+3 \log |\tan x|+\frac{3}{2} \tan ^{2} x+\frac{1}{4} \tan ^{4} x+C \quad\quad\quad\quad[\because t=\tan x] \end{aligned}

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