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How would I show that $$\int_{0}^{\pi /2}\sin^{2n}(t)\, dt = \frac{\pi}{2} \frac{1(3)(5)...(2n-1)}{2(4)(6)(8)...(2n)}$$

using contour integration? My guess is that I would integrate along the unit circle $|z| = 1$ but I'm not sure how to get any further.

John Sweeney
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  • See this answer: http://math.stackexchange.com/questions/476693/using-residue-theorem-to-evaluate-int-0-pi-sin2n-theta-d-theta/476708#476708 I think the question is a duplicate. – Ron Gordon Nov 10 '13 at 21:47

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First note that by symmetry your integral is $1/4$ the integral from $0$ to $2 \pi$. Next, with $z = e^{it}$, $\sin(t) = (z - 1/z)/(2i)$ while $dt = dz/(iz)$. Now find the residue...

Robert Israel
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