Evaluate the following integrals:

Assume $1-\tan ^{2} x=t$

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

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

Substituting $t$ and dt we get

$\Rightarrow \Rightarrow \int \frac{1}{2} \sqrt{t} \mathrm{dt}$

$\Rightarrow \int \frac{1}{2} t^{1} / 2 \cdot d t$

$\Rightarrow \frac{4 t^{\frac{3}{2}}}{6}+c$

But $t=1-\tan ^{2} x$

$\Rightarrow \frac{-2\left(1-\tan ^{2} x\right)^{3 / 2}}{3}+c$