Calculate $\int_{0}^{\pi/2} (\sin(2x))^5 dx$ I want to calculate the following integral: 
$$\int_{0}^{\pi/2} (\sin(2x))^5 dx$$I tried integration by parts but it didn't work for me. Suggestions?

Idea:
Let the integral in the OP be $I$. Split $\sin^3(2x)$ into $\sin^2(2x)$ and $\sin^3(2x)$, and then perform integration by parts. For example, the first step is
\begin{align}
I = \left.\left(\sin^3(2x)\int\sin^2(2x) dx\right)\right|_{0}^{\pi/2} - \int_0^{\pi/2}\left(\int\sin^2(2x) dx\right)\frac{d}{dx}\sin^3(2x)dx
\end{align}
To make things simpler one can also first change the variable $y = 2x$ to get rid of the pesky $2$ in the argument. 
Remark: this is a very common trick to deal with complex integrals involving trigonometric functions due to the the property that they get back to itself (up to $\pm 1$) under twice differentiation, for example
\begin{align}
\frac{d^2}{dx^2}\sin(x) = - \sin(x). 
\end{align}
