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

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

Answers (1)

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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