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?
 A: How about this
\begin{eqnarray}
\mathcal I &=& \int_0^{\pi/2}(\sin 2x)^5 dx=\\
&=&32\int_0^{\pi/2}\sin^5x \cos^5xdx=\\
&=&32\int_0^{\pi/2}\sin^5x(1-\sin^2x)^2\cos x dx=\\
&\stackrel{t=\sin x}{=}&32\int_0^1t^5(1-t^2)^2dt.
\end{eqnarray}
A: hint
Begin by putting $t=2x$, it becomes
$$\frac 12\int_0^\pi\sin(t)^5dt$$
then observe that
$$\sin(t)^5=-(1-\cos(t)^2)^2(\cos(t))'$$
with $u=\cos(t)$, it gives
$$\frac 12\int_{-1}^1(1-u^2)^2du=\int_0^1(1-u^2)^2du$$
$$=1+\frac 15-\frac 23=\frac{8}{15}$$
A: $$\dfrac{d(\sin^nax\cos ax)}{dx}=an\sin^{n-1}ax(1-\sin^2ax)-a\sin^{n+1}x$$
Integrate both sides with respect to $x,$
$$\sin^nax\cos ax=anI_{n+1}-a(n+1)I_{n+1}$$
where $$I_n=\int\sin^max\ dx$$
Set $n+1=5,3,1$ one by one
A: 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}
A: As it is  the integral of a monomial in $\sin 2x$ with an odd exponent, the standard method uses the substitution  $u=\cos 2x$. Indeed, the differential can be rewritten as
$$\sin^5 2x \,\mathrm dx=\sin^4 2x \,\sin2x\,\mathrm dx=(1-u^2)^2\Bigl(-\frac12\,\mathrm du\Bigr),$$ 
so the integral becomes, taking into account the integrand is an even function,
$$-\frac12\int_1^{-1}(1-u^2)^2\,\mathrm d u=\int_0^1(1-2u^2+u^4)\,\mathrm d u.$$
