Solving $\int_0^{\frac{1}{2}} x^{\frac{3}{2}} (1-2x)^{\frac{3}{2}}dx$? How to solve the definite integral $\int_0^{\frac{1}{2}} x^{\frac{3}{2}} (1-2x)^{\frac{3}{2}}~dx$?
Funnily enough, this is actually a problem that showed up during my real analysis course. At first glance, the problem seemed solvable by using knowledge from basic elementary calculus, but it turned out not to be so simple. I initially tried the substitution $u=\sqrt{x}$, but that lead me nowhere... 

Let $u=x^{\frac{1}{2}}$, $du=\frac{1}{2}x^{-\frac{1}{2}}$.
$\int_0^{\frac{1}{2}} x^{\frac{3}{2}} (1-2x)^{\frac{3}{2}}~dx$ = 
  $2\int_0^{\frac{1}{\sqrt{2}}} u^4 (1-2u^2)^{\frac{3}{2}}~du$.

EDIT: I have tried an additional substitution to the above, namely $u= \frac{\sin t}{\sqrt{2}}$.
$$
\begin{align*}
2\int_0^{\frac{1}{\sqrt{2}}} u^4 (1-2u^2)^{\frac{3}{2}}~du&=
2\int_0^{\frac{\pi}{2}} \frac{\sin^4 t}{4} (1-\sin^2 t)^{\frac{3}{2}}\cdot \frac{\cos t}{\sqrt{2}}~dt\\
&=\frac{1}{2\sqrt{2}}\int_0^{\frac{\pi}{2}}\sin^4t\cos^4t~dt.
\end{align*}
$$
Hopefully my calculations aren't wrong...
 A: Note that there $x$ and $(\frac 12 - x)$ is symmetric about $x= \frac 14$. Thus try to substitute $y = \frac 14 -x$. Then the integral is 
$$- \int_{\frac{1}{4}}^{-\frac 14} (\frac 14 -y)^\frac{3}{2} (\frac 12 + 2y)^{\frac 32} dy = 2^{\frac 32} \int_{-\frac 14}^{\frac 14} (\frac 1{16} - y^2)^{\frac 32} dy.$$
From here one can use trigonometric substitution. 
A: $$\int_0^{\frac{1}{2}} x^{\frac{3}{2}} (1-2x)^{\frac{3}{2}}dx$$
Substitute $u = \sqrt{x}$
$$ = 2\int_0^{\frac{1}{\sqrt{2}}} u^4 (1-2u^2)^{\frac{3}{2}}dx$$
Substitute u = $\frac{\sin(s)}{\sqrt{2}}$, $du = \frac{\cos(s)}{\sqrt{2}}$ (this will be valid for $0 < s < \frac{\pi}{2}$)
$$ = \sqrt{2}\int_0^{\frac{\pi}{2}} \frac{1}{4}\sin^4(s)\cos(s)cos^2(s)^{3/2}ds$$
$$ = \sqrt{2}\int_0^{\frac{\pi}{2}} \frac{1}{4}\sin^4(s)\cos^4(s)ds$$
$$ = \frac{1}{2\sqrt{2}}\int_0^{\frac{\pi}{2}} \sin^4(s)\cos^4(s)ds$$
We then apply the following reduction formula with $m=4$ and $n=4$ (which is really messy... I'll leave out a lot of intermediate stuff): 
$$\int \cos^m(s)\sin^m(s)ds = -\frac{\cos^{m+1}(s)\sin^{m+1}(s)}{m+n} + \frac{n-1}{m+n}\int \cos^m(s) \sin^{n-2}(s)ds$$
$$\Rightarrow \frac{3}{16\sqrt{2}}\int_0^{\frac{\pi}{2}} \sin^2(s)\cos^4(s)ds$$
$$=\frac{3}{16\sqrt{2}}\int_0^{\frac{\pi}{2}} (1-\cos^2(s))\cos^4(s)ds$$
$$=\frac{3}{16\sqrt{2}}\int_0^{\frac{\pi}{2}} \big(\cos^4(s) - \cos^6(s)\big)ds$$
$$=\frac{3}{16\sqrt{2}}\int_0^{\frac{\pi}{2}}\cos^4(s)ds - \frac{3}{16\sqrt{2}}\int_0^{\frac{\pi}{2}}\cos^6(s)ds$$
We then apply the following reduction formula for $m=6$:
$$\int \cos^m(s) = \frac{\sin(s)\cos^{m-1}(s)}{m} + \frac{m-1}{m}\int \cos^{m-2}(s)ds$$
$$\Rightarrow \frac{1}{32\sqrt{2}}\int_0^{\frac{\pi}{2}}\cos^4(s)ds$$
Apply the cosine reduction formula again for $m=4$:
$$\Rightarrow \frac{3}{128\sqrt{2}}\int_0^{\frac{\pi}{2}}\cos^2(s)ds$$
We then make the substitution $cos^2(s) = \frac{1}{2}\cos(2s)+\frac{1}{2}$, and after a few very basic calculus and algebra steps arrive at the solution.
$$= \frac{3\pi}{512\sqrt{2}}$$
