Evaluate the following integrals:

Solution:

Put $\tan x=t$

$d(\tan x)=d t$

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

$\Rightarrow \mathrm{d} \mathrm{x}=\frac{\mathrm{dt}}{\sec ^{2} \mathrm{x}}$

We can write $\sec ^{4} x=\sec ^{2} x \cdot \sec ^{2} x$

Now , the question becomes

$\Rightarrow \int \sec ^{2} x \cdot \sec ^{2} x \cdot \tan x \frac{d t}{\sec ^{2} x}$

$\Rightarrow \int \sec ^{2} x \cdot \tan x d t$

$\operatorname{Tan}^{2} x+1=\sec ^{2} x$

$\tan x=t$

$t^{2}+1=\sec ^{2} x$

$\Rightarrow \int\left(\mathrm{t}^{2}+1\right) \mathrm{t} \mathrm{dt}$

$\Rightarrow \int \mathrm{t}^{3} \mathrm{dt}+\int \mathrm{t} \cdot \mathrm{dt}$

$\Rightarrow \frac{t^{4}}{4}+\frac{t^{2}}{2}+c$

But $\mathrm{t}=\tan \mathrm{x}$

$\Rightarrow \frac{\tan ^{4} x}{4}+\frac{\tan ^{2} x}{2}+c$