Answer
$-\dfrac{\sqrt {1-v^2}}{v}-\sin^{-1} v+C$
Work Step by Step
Using the Substitution Method:
$v=\sin \theta \implies dv=\cos \theta \space d \theta$
We integrate the integral as follows:
$I=\int \dfrac{\sqrt {1-\sin^2 \theta})}{(\sin^2 \theta)} \times (\cos \theta \space d \theta) =\int \cot^2 \theta \space d \theta$
Now, $I=-\cot \theta- \theta+C \\= -\dfrac{\sqrt {1-\sin^2 \theta}}{(\sin^2 \theta)}-\theta+C \\=-\dfrac{\sqrt {1-v^2}}{v}-\sin^{-1} v+C$
