Answer
$\dfrac{1}{3} (\dfrac{v}{\sqrt{1-v^2}})^3 +C$
Work Step by Step
Let us consider $v=\sin \theta \implies dv = \cos \theta d\theta$
Now, the given integral can be written as:
$\int\dfrac{\sin^2 \theta \cos \theta d\theta}{\cos^5 \theta} =\dfrac{\tan^3 \theta}{3}+C $
or, $=\dfrac{1}{3} (\dfrac{v}{\sqrt{1-v^2}})^3 +C$
