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.
 A: Hints:
By parts, and taking into account the integrand is an even function
$$-\int_0^1u(-2u)(1-u^2)^{3/2}du=\left.-u\cdot\frac25(1-u^2)^{5/2}\right|_0^1+\frac25\int_0^1(1-u^2)^{5/2}du$$
The first summand on the right is zero, and now for the integral on the right substitute $\;u=\sin t\;$ and get the integral
$$\frac25\int_0^{\pi/2}\cos^6t\,dt=\ldots$$
A: There is not much arduous about evaluating $$\int_0^{\pi}\sin(t)^6\,\mathrm{d}t$$ if you notice that $$\int_0^{\pi}\sin(t)^6\,\mathrm{d}t=\int_0^{\pi}\sin(t)^5\sin(t)\,\mathrm{d}t$$ $$=\int_0^{\pi}5\sin(t)^4\cos(t)^2\,\mathrm{d}t-\sin(\pi)^5\cos(\pi)+\sin(0)^5\cos(0)=\int_0^{\pi}5\sin(t)^4[1-\sin(t)^2]\,\mathrm{d}t$$ $$=5\int_0^{\pi}\sin(t)^4\,\mathrm{d}t-5\int_0^{\pi}\sin(t)^6\,\mathrm{d}t,$$  which is equivalent to $$\int_0^{\pi}\sin(t)^6\,\mathrm{d}t=\frac56\int_0^{\pi}\sin(t)^4\,\mathrm{d}t,$$ and that this can be further simplified recursively, so that $$\int_0^{\pi}\sin(t)^6\,\mathrm{d}t=\frac56\int_0^{\pi}\sin(t)^4\,\mathrm{d}t=\frac56\frac34\int_0^{\pi}\sin(t)^2\,\mathrm{d}t.$$
A: 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
$$
au^{-n} + b = x^s, \text{ where }s\text{ is a denominator of }p, \text{ i.e.,} s = 2.
$$
$$
\begin{aligned}
\int_{-1}^1{u^2(1 - u^2)^\frac{3}{2}du} &= 2\int_{0}^1{u^2(1 - u^2)^\frac{3}{2}du} = \\
&= \left|
\begin{aligned}
x^2 = u^{-2} - 1 &\Leftrightarrow u^2 = \frac{1}{x^2+1} \\
2udu = -\frac{2xdx}{(x^2+1)^2} &\Leftrightarrow du = -\frac{x}{(x^2+1)^2}\sqrt{x^2+1}dx \\
u = 0 &\Leftrightarrow x = +\infty \\
u = 1 &\Leftrightarrow x = 0 \\
\end{aligned}
\right| = \\
&= 2\int_0^{+\infty}\frac{1}{x^2+1}\left(\frac{x^2}{x^2+1}\right)^\frac{3}{2}\frac{x}{(x^2+1)^2}\sqrt{x^2+1}dx = \\
&= 2\int_0^{+\infty}\frac{x^4}{(x^2+1)^4}dx = 
-\frac{1}{3}\int_0^{+\infty}x^3d\left\{\frac{1}{(x^2+1)^3}\right\} = |\text{integrating by parts}| = \\
&= -\frac{1}{3}\left(\underbrace{\left.x^3\frac{1}{(x^2+1)^3}\right|_0^{+\infty}}_{=0}-\int_0^{+\infty}\frac{1}{(x^2+1)^3}3x^2dx\right) = \\
&= \int_0^{+\infty}x^2\frac{1}{(x^2+1)^3}dx = -\frac{1}{4}\int_0^{+\infty}xd\left\{\frac{1}{(x^2+1)^2}\right\} =|\text{intgrating by parts}| = \\
&= -\frac{1}{4}\left(\underbrace{\left.x\frac{1}{(x^2+1)^2}\right|_0^{+\infty}}_{=0}-\int_0^{+\infty}\frac{1}{(x^2+1)^2}dx\right) = \\
&= \frac{1}{4}\int_0^{+\infty}\frac{1}{(x^2+1)^2}dx = \frac{1}{4}\int_0^{+\infty}\left(\frac{x^2+1}{(x^2+1)^2}-\frac{x^2}{(x^2+1)^2}\right)dx = \\
&= \frac{1}{4}\underbrace{\int_0^{+\infty}\frac{1}{x^2+1}dx}_{= \left.\arctan(x)\right|_0^{+\infty} = \frac{\pi}{2}}-\frac{1}{4}\int_0^{+\infty}\frac{x^2}{(x^2+1)^2}dx= \\
&= \frac{\pi}{8}+\frac{1}{8}\int_0^{+\infty}xd\left\{\frac{1}{x^2+1}\right\} = |\text{integrating by parts}| = \\
&=\frac{\pi}{8}+\frac{1}{8}\left(\underbrace{\left.x\frac{1}{x^2+1}\right|_0^{+\infty}}_{=0}-\int_0^{+\infty}\frac{1}{x^2+1}dx\right) = \frac{\pi}{8} -\frac{1}{8}\underbrace{\int_0^{+\infty}\frac{1}{x^2+1}}_{=\frac{\pi}{2}} = \frac{\pi}{8}-\frac{\pi}{16} = \frac{\pi}{16}.
\end{aligned}
$$
A: Make the change of variables $u\mapsto \sqrt{u}$ so that your integral becomes
$$\int_0^1 u^{1/2}(1-u)^{3/2}\,du,$$
which, by appealing to the Beta function turns into
$$\beta(3/2, 5/2)=\frac{1}{6}\Gamma(3/2)\Gamma(5/2)=\pi/16.$$
