Calculate $\int^1_{-1}\sqrt{(1-x^2)^3} dx$ using Integration by parts or substitution technique How can I calculate $\int^1_{-1}\sqrt{(1-x^2)^3} dx$?.
I'm trying to solve it by using substitutions or integrate by parts but none of those leads me to the solution. 
My attempt (with the help of a friend):
first since $(1-x^2)\ge 0$ than $\sqrt{(1-x^2)^3}$ could be written as $(1-x^2)^{2/3}$.
Now, $u=(1-x^2)^{2/3}, u' = -3x\sqrt{1-x^2}$. $v'=1, v =x$
and so $\int^1_{-1}\sqrt{(1-x^2)^3} dx = [x(1-x^2)^{3/2}]^1_{-1}+\int^1_{-1}3x^2\sqrt{1-x^2} dx$.
I don't even know if it's good, and if it is, I'm completely clueless at how to find $\int^1_{-1}3x^2\sqrt{1-x^2} dx$

"$1-x^2$" should immediately make you think about the substitution $x=\sin u$, as $1-x^2 = 1-\sin^2 u = \cos^2 u$. So let $x = \sin u$, then $dx = \cos u \,du$ and note that $\sin (-\tfrac{\pi}{2}) = -1$ and $\sin (\tfrac{\pi}{2}) = 1$, so
\begin{align}
    \int_{-1}^{1} (1-x^2)^{3 / 2}\,dx
&= \int_{-\pi/2}^{\pi/2} (1-\sin^2 u)^{3 / 2} \cos u\, du
= \int_{-\pi/2}^{\pi/2} (\cos^2 u)^{3 / 2} \cos u\, du\\
&= \int_{-\pi/2}^{\pi/2} \cos^4 u\, du
= \int_{-\pi/2}^{\pi/2} (\cos^2 u)^2\, du\\
&= \int_{-\pi/2}^{\pi/2} \left(\tfrac{1}{2} + \tfrac{1}{2} \cos(2u)\right)^2\, du\\
&= \int_{-\pi/2}^{\pi/2} \tfrac{1}{4} + \tfrac{1}{2}\cos(2u) + \tfrac{1}{4}\cos^2(2u)\; du\\
&= \int_{-\pi/2}^{\pi/2} \tfrac{1}{4} + \tfrac{1}{2}\cos(2u) + \tfrac{1}{4}\left(\tfrac{1}{2}+\tfrac{1}{2}\cos (4u) \right)\; du\\
&= \int_{-\pi/2}^{\pi/2} \tfrac{3}{8} + \tfrac{1}{2}\cos(2u) + \tfrac{1}{8}\cos (4u)\; du\\
\end{align}
Now just use a bit of substitution to finish up.
