The value of $\int \frac{1}{\sqrt{\sin ^3 x \cos ^5 x}} d x$, is

$\frac{-2}{\sqrt{\tan x}}+\frac{2}{3}(\tan x)^{3 / 2}+C$ 

We have,

$I =\int \frac{1}{\sqrt{\sin ^3 x \cos ^5 x}} d x$

$\Rightarrow I =\int \frac{1}{\sin ^{3 / 2} x \cos ^{5 / 2} x} d x$

$\Rightarrow I =\int \frac{\sec ^4 x}{\tan ^{3 / 2} x} d x$         [Dividing $N^r$ and $D^r$ by $\cos ^2 x$]

$\Rightarrow I=\int \frac{\left(1+\tan ^2 x\right)}{\tan ^{3 / 2} x} d(\tan x)$

$\Rightarrow I=\int\left\{(\tan x)^{-3 / 2}+(\tan x)^{1 / 2}\right\} d(\tan x)$

$\Rightarrow I=-2(\tan x)^{-1 / 2}+\frac{2}{3}(\tan x)^{3 / 2}+C$

$\Rightarrow I=\frac{-2}{\sqrt{\tan x}}+\frac{2}{3}(\tan x)^{3 / 2}+C$