Answer
$-\dfrac{1}{3}(\dfrac{\sqrt{1-x^2}}{x})^3+C$
Work Step by Step
Let us consider $x = \sin \theta $
and $dx= \cos \theta d \theta $
Now, the given integral becomes:
$\int \dfrac{\cos \theta}{\sin ^4 \theta} (\cos \theta) d\theta=\int csc^2 \theta (\cot^2 \theta) d \theta $
Now, integrate:
$-\dfrac{\cot^3 \theta}{3} +C=-\dfrac{1}{3}(\dfrac{\sqrt{1-x^2}}{x})^3+C$
