Evaluate the following integrals:

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

$\Rightarrow I=\int \sqrt{\tan x} \sec ^{2} x \sec ^{2} x d x$

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

$\Rightarrow I=\int\left(\tan ^{\frac{1}{2}} x+\tan ^{\frac{5}{2}} x\right) \sec ^{2} x d x$

Let $\tan x=t$, then

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

$\Rightarrow I=\int\left(t^{\frac{1}{2}}+t^{\frac{5}{2}}\right) d t$

$\Rightarrow I=\frac{2}{3} t \overline{2}+\frac{2}{7} t \frac{7}{2}+c$

$\Rightarrow I=\frac{2}{3} \tan \frac{3}{2} x+\frac{2}{7} \tan ^{\frac{7}{2}} x+c$

Therefore, $\int \sqrt{\tan x} \sec ^{4} x d x=\frac{2}{3} \tan ^{\frac{3}{2}} x+\frac{2}{7} \tan ^{\frac{7}{2}} x+c$