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I'm trying to solve the integral $\int^{2\pi}_0\cos^2(\theta) \sin^2(\theta) d\theta $.

So far I've used $\cos\theta=\frac{1}{2}(z+z^{-1})$, $\sin\theta=-\frac{1}{2}i(z-z^{-1})$ and $d\theta=-iz^{-1}dz$ to get the contour integral $$\int^{2\pi}_0\cos^2(\theta) \sin^2(\theta) d\theta = \frac{i}{8}\int\frac{z^3}{(z+1)^4(z-1)^4} dz$$ (It's a curve integral, but I don't know how to write that)

Therefore there are poles in z=1 and z=-1, both of order 4.

Problem is that to actually calculate the integral I need to know the residues to use the residue theorem and everytime I try toget the residues, I get 0. No matter which pole I'm trying to calculate.

To calculate the residues, I'm using the formula $R(z_0)=\lim_{z->z_0}\left(\frac{1}{(m-1)!}\frac{d^{m--1}}{dz^{m-1}}[(z-z_0)^mf(z)]\right)$, where m is the order pole.

Can anyone see what I'm doing wrong?

Nana
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    I do not get the complex integral you state when substituting. Numertaor and denominator seem interchanged? Also in cos there is a + between the terms – H. H. Rugh Aug 24 '16 at 07:29
  • Just do the substitution one more time – Zau Aug 24 '16 at 08:11

2 Answers2

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This is a simpler way: $$\int^{2\pi}_0\cos^2(\theta) \sin^2(\theta) d\theta=\frac{1}{4}\int^{2\pi}_0\sin^2(2\theta) d\theta= \frac{1}{8}\int^{2\pi}_0(1-\cos(4\theta)) d\theta=\frac{2\pi}{8}-0=\frac{\pi}{4}.$$ As regards the complex approach, you should have $$\int^{2\pi}_0\cos^2(\theta) \sin^2(\theta) d\theta = -\frac{1}{16}\int_{|z|=1}(z+1/z)^2(z-1/z)^2 \frac{dz}{iz}\\=-\frac{2\pi i}{16i}\mbox{Res}\left(z^3-\frac{2}{z}+\frac{1}{z^5}, 0\right)=\frac{\pi}{4}.$$

Robert Z
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  • I'm studying for an exam where it's important I make a use of contour integration with residues and so on. So even though it's correct, I'm afraid I'm looking for a way to do it that way even though this is simplier – Nana Aug 24 '16 at 08:01
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    @Nana I understand. I answered also for the complex approach. – Robert Z Aug 24 '16 at 08:04
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Another simple way:

Out of symmetry:

$$\int^{2\pi}_0\cos^2(\theta) \sin^2(\theta) d\theta = 4 \int^{\pi/2}_0\cos^2(\theta) \sin^2(\theta) d\theta $$

Recall that $2 \int^{\pi/2}_0\cos^2(\theta) \sin^2(\theta) d\theta = B(\frac{3}{2},\frac{3}{2}) = \frac{\Gamma(\frac{3}{2})^2}{2} = \frac{\frac{1}{4}\Gamma(\frac{1}{2})^2}{2} = \frac{\pi}{8} $

So $$\int^{2\pi}_0\cos^2(\theta) \sin^2(\theta) d\theta = 2\frac{\pi}{8} = \frac{\pi}{4}$$

Zau
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