Evaluate the following integrals:

$\int \tan ^{4} x d x$

We can write above integral as:

$\int \tan ^{4} x d x=\int\left(\tan ^{2} x\right)\left(\tan ^{2} x\right) d x \cdots\left(\right.$ Splitting $\left.\tan ^{4} x\right)$

$=\int\left(\sec ^{2} x-1\right) \tan ^{2} x d x\left(U \operatorname{sing} \tan ^{2} x=\sec ^{2} x-1\right)$

Considering integral (1)

Let $u=\tan x$

$d u=\sec ^{2} x d x$

Substituting values we get,

$\int \sec ^{2} x\left(\tan ^{2} x\right) d x=\int u^{2} d u=\frac{u^{3}}{3}+C$

Substituting value of u we get,

$\int \sec ^{2} x\left(\tan ^{2} x\right) d x=\frac{\tan ^{3} x}{3}+C$

Considering integral (2)

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

$=\int\left(\sec ^{2} x\right) d x-\int 1 d x$

$=\tan x-x+C$

$\therefore$ integral becomes,

$\int \sec ^{2} x\left(\tan ^{2} x\right) d x-\int\left(\tan ^{2} x\right) d x=\frac{\tan ^{3} x}{3}+C-(\tan x-x+C)$

$=\frac{\tan ^{2} x}{3}-\tan x+x+C[\because C+C$ is a constant $]$

$\therefore \int \tan ^{4} x d x=\frac{\tan ^{3} x}{3}-\tan x+x+C$