Solution:

Assume $I=\int \sqrt{\tan x} \sec ^{4} x d x$
We can write the above equation as
$\Rightarrow I=\int \sqrt{\tan x} \sec ^{2} x \sec ^{2} x d x$
Now, taking common
$\begin{array}{l}
\Rightarrow I=\int \sqrt{\tan x}\left(1+\tan ^{2} x\right) \sec ^{2} x d x \\
\Rightarrow I=\int\left(\tan ^{\frac{1}{2}} x+\tan ^{\frac{5}{2}} x\right) \sec ^{2} x d x
\end{array}$
Assume $\tan \mathrm{x}=\mathrm{t}$, therefore
$\begin{array}{l}
\Rightarrow \sec ^{2} x d x=d t \\
\Rightarrow I=\int\left(t^{\frac{1}{2}}+t^{\frac{5}{2}}\right) d t
\end{array}$
On integrating we obtain
$\Rightarrow \mathrm{I}=\frac{2}{3} \mathrm{t}^{\frac{3}{2}}+\frac{2}{7} \mathrm{t}^{\frac{7}{2}}+\mathrm{c}$
On substituting the value of $\mathrm{t}$, we get
$\Rightarrow I=\frac{2}{3} \tan ^{\frac{3}{2}} x+\frac{2}{7} \tan ^{\frac{7}{2}} x+c$
Hence, $\int \sqrt{\tan x} \sec ^{4} x d x=\frac{2}{3} \tan ^{\frac{3}{2}} x+\frac{2}{7} \tan ^{\frac{7}{2}} x+c$