#### Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.16 question 15

$\frac{1}{6} \log \left|\frac{1+\tan ^{3} x}{1-\tan ^{3} x}\right|+C$

Hint:

Use substitution method as well as special integration formula to solve this type of problem

Given:

$\int \frac{\tan ^{2} x \sec ^{2} x}{1-\tan ^{6} x} d x$

Solution:

$Let\: \: I=\int \frac{\tan ^{2} x \sec ^{2} x}{1-\tan ^{6} x} d x$

$=\int \frac{\tan ^{2} x \sec ^{2} x}{1-(\tan ^{3} x)^{2}} d x$

\begin{aligned} &\text { Put } \tan ^{3} x=t \Rightarrow 3 \tan ^{2} x \cdot \sec ^{2} x d x=d t \\ &\text { Then } I=\int \frac{1}{1-t^{2}} \cdot \frac{d t}{3} \\ &=\frac{1}{3} \int \frac{1}{1^{2}-t^{2}} d t \\ &=\frac{1}{3} \cdot \frac{1}{2 \times 1} \log \left|\frac{1+t}{1-t}\right|+C \quad\left[\because \int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\right] \\ &=\frac{1}{6} \log \left|\frac{1+\tan ^{3} x}{1-\tan ^{3} x}\right|+C \quad\left[\because t=\tan ^{3} x\right] \end{aligned}