Evaluate the following integrals:

above equation can be solve by using one formula that is $\left(i+\tan ^{2} x=\sec ^{2} x\right)$

$I=\int \sec ^{4} x d x$

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

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

$=\int \sec ^{2} x d x+\int \sec ^{2} x \tan ^{2} x d x$

Put $\tan x=t$ in above equation so that $\sec ^{2} x d x=d t$

$I=\tan x+\int t^{2} d t=\tan x+\frac{t^{3}}{3}$

$=\tan x+\frac{\tan ^{3} x}{3}$