Evaluating $\int_{-1}^{1} u^2(1-u^2)^{3/2}du$ The integral I'm having difficulty with solving is:
$I = \int_{-1}^{1} u^2(1-u^2)^{3/2}du$
I arrived at this integral in trying to solve
$\int_{0}^{\pi} sin^4(x)cos^{2}(x)dx$
by making the substitution $u=cos(x)$
I attempted integration by parts as follows:
$F=u, \: F'=1 \\G=-\frac{1}{5}(1-u^2)^{\frac{5}{2}}, \: G'=u(1-u^2)^{\frac{3}{2}},$
Such that
$I = FG|_{-1}^{1} - \int F' G = \frac{1}{5}\int_{-1}^{1}(1-u^2)^{\frac{5}{2}} $
After which I think the obvious thing to do is to again make a trig substitution $u=cos(x)$, which leads to:
$I= \frac{1}{5} \int_{\pi}^{0} sin^{6}(x) dx$
This seems like an arduous route to solving this integral. Can anyone come up with an easier one?
Thanks for your help in advance.

The integral expression looks like differnetial binom
$$
I = \int{u^m(a + bu^n)^pdu}.
$$
Comparing that to the initial integral,
$$
m = 2, n = 2, p = \frac{3}{2}, a = 1, b = -1.
$$
$$
p + \frac{m+1}{n} = \frac{3}{2} + \frac{2+1}{2} = 3
$$
is an integer. Then the following substitution is used
$$
au^{-n} + b = x^s, \text{ where }s\text{ is a denominator of }p, \text{ i.e.,} s = 2.
$$
$$
\begin{aligned}
\int_{-1}^1{u^2(1 - u^2)^\frac{3}{2}du} &= 2\int_{0}^1{u^2(1 - u^2)^\frac{3}{2}du} = \\
&= \left|
\begin{aligned}
x^2 = u^{-2} - 1 &\Leftrightarrow u^2 = \frac{1}{x^2+1} \\
2udu = -\frac{2xdx}{(x^2+1)^2} &\Leftrightarrow du = -\frac{x}{(x^2+1)^2}\sqrt{x^2+1}dx \\
u = 0 &\Leftrightarrow x = +\infty \\
u = 1 &\Leftrightarrow x = 0 \\
\end{aligned}
\right| = \\
&= 2\int_0^{+\infty}\frac{1}{x^2+1}\left(\frac{x^2}{x^2+1}\right)^\frac{3}{2}\frac{x}{(x^2+1)^2}\sqrt{x^2+1}dx = \\
&= 2\int_0^{+\infty}\frac{x^4}{(x^2+1)^4}dx = 
-\frac{1}{3}\int_0^{+\infty}x^3d\left\{\frac{1}{(x^2+1)^3}\right\} = |\text{integrating by parts}| = \\
&= -\frac{1}{3}\left(\underbrace{\left.x^3\frac{1}{(x^2+1)^3}\right|_0^{+\infty}}_{=0}-\int_0^{+\infty}\frac{1}{(x^2+1)^3}3x^2dx\right) = \\
&= \int_0^{+\infty}x^2\frac{1}{(x^2+1)^3}dx = -\frac{1}{4}\int_0^{+\infty}xd\left\{\frac{1}{(x^2+1)^2}\right\} =|\text{intgrating by parts}| = \\
&= -\frac{1}{4}\left(\underbrace{\left.x\frac{1}{(x^2+1)^2}\right|_0^{+\infty}}_{=0}-\int_0^{+\infty}\frac{1}{(x^2+1)^2}dx\right) = \\
&= \frac{1}{4}\int_0^{+\infty}\frac{1}{(x^2+1)^2}dx = \frac{1}{4}\int_0^{+\infty}\left(\frac{x^2+1}{(x^2+1)^2}-\frac{x^2}{(x^2+1)^2}\right)dx = \\
&= \frac{1}{4}\underbrace{\int_0^{+\infty}\frac{1}{x^2+1}dx}_{= \left.\arctan(x)\right|_0^{+\infty} = \frac{\pi}{2}}-\frac{1}{4}\int_0^{+\infty}\frac{x^2}{(x^2+1)^2}dx= \\
&= \frac{\pi}{8}+\frac{1}{8}\int_0^{+\infty}xd\left\{\frac{1}{x^2+1}\right\} = |\text{integrating by parts}| = \\
&=\frac{\pi}{8}+\frac{1}{8}\left(\underbrace{\left.x\frac{1}{x^2+1}\right|_0^{+\infty}}_{=0}-\int_0^{+\infty}\frac{1}{x^2+1}dx\right) = \frac{\pi}{8} -\frac{1}{8}\underbrace{\int_0^{+\infty}\frac{1}{x^2+1}}_{=\frac{\pi}{2}} = \frac{\pi}{8}-\frac{\pi}{16} = \frac{\pi}{16}.
\end{aligned}
$$
