Question:
$\int \tan ^{2} x \sec ^{4} x d x$
Solution:
Let $\mathrm{I}=\int \tan ^{2} x \cdot \sec ^{4} x d x$
$=\int \tan ^{2} x \sec ^{2} x \cdot \sec ^{2} x d x=\int \tan ^{2} x\left(1+\tan ^{2} x\right) \cdot \sec ^{2} x d x$
Putting, $\tan x=t, \Rightarrow \sec ^{2} x d x=d t$
$I=\int t^{2}\left(1+t^{2}\right) d t=\int\left(t^{2}+t^{4}\right) d t=\int t^{2} d t+\int t^{4} d t$
$=\frac{1}{3} t^{3}+\frac{1}{5} t^{5}=\frac{1}{3} \tan ^{3} x+\frac{1}{5} \tan ^{5} x+C$
Therefore,
$\mathrm{I}=\frac{1}{3} \tan ^{3} x+\frac{1}{5} \tan ^{5} x+\mathrm{C}$