Easiest way to prove that $\int_{\pi/6}^{\pi/2} \sin(2x)^3\cos(3x)^2 \mathrm{d}x=\left(3/4\right)^4$ I have been trying to evaluate this integral a few times. And my best attempt has been to rewrite is as a sum of linear combination of sine and cosine terms. Alas, this takes a couple of handwritten pages to accomplish. Is there any easier/faster/neater way to evaluate? 
$$ \int_{\pi/6}^{\pi/2} \sin(2x)^3\cos(3x)^2\,\mathrm{d}x=\left(\frac{3}{4}\right)^4 $$ 
Thanks in advance =)
 A: Write 
$$ \sin 2x = \frac{e^{2ix}-e^{-2ix}}{2i}$$
and
$$ \cos 3x = \frac{e^{3ix}+e^{-3ix}}{2},$$
cube the first identity, square the second, and expand.
A: The way you describe is sensible. The notation is a little misleading, you probably intend
$$\int_{\pi/6}^{\pi/2} (\sin 2x)^3 (\cos 3x)^2\,dx.$$ If that is so, the answer is correct.
Unpleasantness is difficult to avoid here. A way to express as a linear combination is to start by using $\cos 2u=2\cos^2 u-1=1-2\sin^2 u$. Thus our integral becomes
$$\frac{1}{4}\int_{\pi/6}^{\pi/2}(\sin 2x)(1-\cos 4x)(1+\cos 6x).$$ 
Now continue. To save some writing, let $u=2x$. Then we want
$$\frac{1}{8}\int_{\pi/3}^\pi(\sin u)(1-\cos 2u)(1+\cos 3u)\,du.$$
using the formulas for expressing a product of sines and/or cosines as a sum of sines and cosines, which you probably already used. Easier by hand than with LaTeX!  
A: $$\int_{\pi/6}^{\pi/2} (\sin 2x)^3 (\cos 3x)^2\,dx$$ 
Since $(\sin 2x)^2=1-\cos 4x$ and $(\cos 3x)^2=1+\cos 6x$
$$=
\frac{1}{4}\int_{\pi/6}^{\pi/2}(\sin 2x)(1-\cos 4x)(1+\cos 6x)$$
Substitute $u=2x$,
$$=\frac{1}{8}\int_{\pi/3}^\pi(\sin u)(1-\cos 2u)(1+\cos 3u)\,du$$
Use the formula $\sin u=\frac{e^{iu}-e^{-iu}}{2i}$ and $\cos u=\frac{e^{iu}+e^{-iu}}{2}$,
$$= \frac{1}{8}\int_{\pi/3}^\pi \frac{e^{iu}-e^{-iu}}{2i}(1- \frac{e^{2iu}+e^{-2iu}}{2} 
)(1+\frac{e^{3iu}+e^{-3iu}}{2})du$$
Expand,
$$= \frac{1}{64i}\int_{\pi/3}^\pi( - (e^{6iu}-e^{-6iu})+3(e^{4iu}-e^{-4iu})-2(e^{3iu}-e^{-3iu})-3(e^{2iu}-e^{-2iu})+6(e^{iu}-e^{-iu}) )du$$
Use the formula $\sin u=\frac{e^{iu}-e^{-iu}}{2i}$, go back to $\sin$
$$= \frac{1}{32}\int_{\pi/3}^\pi(-(\sin 6u)+3(\sin 4u)-2(\sin 3u)-3(\sin 2u)+6(\sin u))du$$
Do the integral, in total five parts,
$$= \frac{1}{32}(\cos 6u/6-3\cos 4u/4+2\cos 3u/3+3\cos 2u/2-6\cos u)|_{\pi/3}^\pi$$
Calculate the value for each of the five parts,
$$= \frac{1}{32}(0-9/8+0+9/4+9)=\frac{3^4}{4^4}$$
