$\int \tan ^{5} x d x$
$\int \tan ^{5} x d x$
We can write above integral as:
$\int \tan ^{5} x d x=\int\left(\tan ^{3} x\right)\left(\tan ^{2} x\right) d x \cdots$ (Splitting $\tan ^{5} \mathrm{x}$ )
$=\int \tan ^{3} x\left(\sec ^{2} x-1\right) d x$ (Using $\tan ^{2} x=\sec ^{2} x-1$ )
$=\int \sec ^{2} x\left(\tan ^{3} x\right) d x-\int\left(\tan ^{3} x\right) d x$
$=\int \sec ^{2} x\left(\tan ^{3} x\right) d x-\int\left(\tan ^{2} x\right)(\tan x) d x-\left(\right.$ Splitting $\left.\tan ^{3} \mathrm{x}\right)$
$=\int \sec ^{2} x\left(\tan ^{3} x\right) d x-\int\left(\sec ^{2} x-1\right)(\tan x) d x$
(Using $\tan ^{2} x=\sec ^{2} x-1$ )
Considering integral (1)
Let $u=\tan x$
$d u=\sec ^{2} x d x$
Substituting values we get,
$\int \sec ^{2} x\left(\tan ^{3} x\right) d x=\int u^{3} d u=\frac{u^{4}}{4}+C$
Substituting value of u we get,
$\int \sec ^{2} x\left(\tan ^{3} x\right) d x=\frac{\tan ^{4} x}{4}+C$
Considering integral (2)
Let $t=\tan x$
$d t=\sec ^{2} x d x$
Substituting values we get,
$\int \sec ^{2} x(\tan x) d x=\int t d t=\frac{t^{2}}{2}+C$
Substituting value of $t$ we get,
$\int \sec ^{2} x(\tan x) d x=\frac{\tan ^{2} x}{2}+C$
Considering integral (3)
$\int(\tan x) d x=-\log |\cos x|\left[\because \int \tan x d x=-\log |\cos x|+C\right]$
$\therefore$ integral becomes,
$\int \sec ^{2} x\left(\tan ^{3} x\right) d x-\int \sec ^{2} x(\tan x) d x-\int(\tan x) d x$
$=\frac{\tan ^{4} x}{4}+C-\left(\frac{\tan ^{2} x}{2}+C\right)-(-\log |\cos x|)$
$=\left(\frac{\tan ^{4} x}{4}\right)+\left(\frac{\tan ^{2} x}{2}\right)+(\log |\cos x|)+C[\because C+C+C$ is a constant $]$
$\therefore \int \tan ^{5} x d x=\left(\frac{\tan ^{4} x}{4}\right)+\left(\frac{\tan ^{2} x}{2}\right)+(\log |\cos x|)+C$