Answer
$-\dfrac{1}{5}(\dfrac{\sqrt{1-x^2}}{x})^5+C$
Work Step by Step
Let us consider $x = \sin \theta $
and $dx= \cos \theta d \theta $ and $\cos \theta=\sqrt {1-x^2} \implies \cos^3 \theta=(\sqrt {1-x^2})^3$
Now, the given integral becomes:
$\int\dfrac{(\cos^3 \theta)(\cos \theta)}{\sin ^6 \theta} =\int \dfrac{\cos^4 \theta}{\sin ^6 \theta} =\int csc^2 \theta (\cot^4 \theta) d \theta $
Now, integrate and plug in $\dfrac{\cos \theta }{\sin \theta }=\cot \theta =\dfrac{\sqrt{1-x^2}}{x}$
$-\dfrac{\cot^5 \theta}{5} +C=-\dfrac{1}{5}(\dfrac{\sqrt{1-x^2}}{x})^5+C$
