# 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?

$$\int_{\pi/6}^{\pi/2} \sin(2x)^3\cos(3x)^2\,\mathrm{d}x=\left(\frac{3}{4}\right)^4$$

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The easiest way, huh? What could be more subjective. –  Gigili Jun 13 '12 at 22:35

$$\int_{\pi/6}^{\pi/2} (\sin 2x)^3 (\cos 3x)^2\,dx$$ 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, $$= \frac{1}{32}(0-9/8+0+9/4+9)=\frac{3^4}{4^4}$$

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@Peter Phipps: Thanks, I edited the answer acoording to your advice. –  caozhu Jun 15 '12 at 7:08
+1, nice answer. –  Peter Phipps Jun 15 '12 at 12:22

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.

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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!

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+1, really good answer. –  Iyengar Jun 14 '12 at 17:32