# Thread: MIT INTEGRATION Bee problem

1. ## MIT INTEGRATION Bee problem



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

2. ## 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?

3. ## 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?

### mit integration bee

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