# MIT INTEGRATION Bee problem

• Sep 8th 2012, 12:49 AM
kyb03316
MIT INTEGRATION Bee problem
http://s0.wp.com/latex.php?latex=%5C...&fg=545454&s=0

... someone solve this problem by complex integration, but i don't know it;

• Sep 8th 2012, 01:59 AM
Vlasev
Re: MIT INTEGRATION Bee problem
Since you mention complex integration, try the following. The complex sine is defined as

$\sin w = \frac{1}{2i}(e^{iw}-e^{-iw})$

$\sin(nx)\sin^m x = \frac{e^{inx}-e^{-inx}}{2i}\frac{(e^{ix}-e^{-ix})^m}{(2i)^m}$

Let $z = e^{ix}$. To find the integral we also need to see how $dx$ transforms. $dz = d(e^{ix}) = ie^{ix} dx = iz dx$, so $dx = dz/(iz)$ and so the integrand becomes

$2^{-m-1}i^{-m-2}(z^n-z^{-n})(z-z^{-1})^m$

In your situation you have $n= m +2$. Does this help?
• Sep 8th 2012, 02:13 AM
Vlasev
Re: MIT INTEGRATION Bee problem
Actually, here is a much easier way to perform the integration. Your integrand is of the form

$\sin((m+1)x)\sin^{m-1}x$

where $m$ is a positive integer. You have that

$= (\sin^{m-1}x)\sin(mx+x) = (\sin^{m-1}x)(\cos(mx)\sin x - \cos x\sin(mx))$

Expanding this, we get

$\cos(mx)\sin^m x - \sin(mx)\sin^{m-1}x \cos x$

Now, can you recognize this as the derivative of something?