Evaluate the following integrals:

Solution:

$\int \tan ^{5} x d x$

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 $]$

$\therefore \int \tan ^{5} x d x=\left(\frac{\tan ^{4} x}{4}\right)+\left(\frac{\tan ^{2} x}{2}\right)+(\log |\cos x|)+C$