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

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

 A: 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, $$I = -\int t^4 dt = -\frac{t^5}{5} + c = -\frac{\cot^5(\theta)}{5} + c = -\frac15 \left( \frac{\sqrt{1-x^2}}{x} \right)^5 + c$$
A: Integral,
$$
\begin{align*}
I &= \int \frac{\sqrt{(1-x^2)^{3}}}{x^6} dx 
\\ &= \int \left( \frac{1-x^2}{x^2} \right)^{3/2} \frac{1}{x^3}dx
\\ &= \int \left(\frac{1}{x^2}-1 \right)^{3/2} \cdot \frac{1}{x^3} dx
\end{align*}
$$
Make the substitution: $z= \frac{1}{x^2}-1$, so that $dz=\frac{-2}{x^3} ~dx$.
$$
\begin{align*}
\text{Integral}
&= -\frac{1}{2}\int z^{3/2} dz 
\\ &= -\frac{1}{2} \cdot \frac{2}{5}z^{5/2}+C
\\ &= -\frac{1}{5} \left(\frac{1}{x^2} - 1 \right)^{5/2}+C
\\ &= - \frac{1}{5} \left(\frac{\sqrt{1-x^2}}{x} \right)^5+C.
\end{align*}
$$
A: Hint: Do the substitution $x = \sin(\alpha)$, $\alpha \in [-\frac{\pi}{2},\frac{\pi}{2}]$. Then you get
$$\int \frac{\cos^4(\alpha)}{\sin^6(\alpha)} d\alpha.$$
Now consider the following identities:
$$D\left(\frac{\cos^3(\alpha)}{\sin^5(\alpha)}\right) = \frac{-3\cos^2(\alpha)\sin^6(\alpha) - 5\sin^4(\alpha)\cos^4(\alpha)}{\sin^10(\alpha)} = -3\frac{\cos^2(\alpha)}{\sin^4(\alpha)} - 5 \frac{\cos^4(\alpha)}{\sin^6(\alpha)}$$
$$D\left(\frac{\cos(\alpha)}{\sin^3(\alpha)}\right) = \frac{-\sin^4(\alpha)-3\sin^2(\alpha)\cos^2(\alpha)}{\sin^6(\alpha)} = -\frac{1}{\sin^2(\alpha)} - 3 \frac{\cos^2(\alpha)}{\sin^4(\alpha)}$$
By using these it boils down to being able to integrate $\frac{1}{\sin^2(\alpha)}$.
