Integrate $\int\frac{1}{x^6} \sqrt{(1-x^2)^3} ~ dx$ How to integrate the following? 

$$\int\frac{\sqrt{(1-x^2)^3}}{x^6} \;dx .$$


Let $x = \sin(\theta)$. We then get $dx = \cos(\theta) d \theta$. Hence, $$I= \int \frac{\cos^3(\theta)}{\sin^6(\theta)} \cos(\theta) d \theta = \int \cot^4(\theta) cosec^2(\theta) d \theta$$
Let $\cot(\theta) = t$, then $-cosec^2(\theta) d \theta = dt$. Hence, $$I = -\int t^4 dt = -\frac{t^5}{5} + c = -\frac{\cot^5(\theta)}{5} + c = -\frac15 \left( \frac{\sqrt{1-x^2}}{x} \right)^5 + c$$
