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Evaluating $\displaystyle \int_{0}^{2\pi}\frac{1}{4\cos^2 t + 9\sin^2 t} dt$

Can someone show me a way using Complex Analysis methods? This is what I tried:

Let $C$ be the contour defined by $C(t)= 2\cos t + 3i\sin t$, $0 \leq t \leq 2\pi$.

Considering the integral

$$I= \int_{C}\frac{dz}{z}= 2\pi i$$
by Cauchy's Integral Formula, then also
$$I = \int_{C}\frac{\overline{z}}{|z|^2}= \int_{0}^{2\pi}\frac{2\cos t - 3i\sin t}{4\cos^2 t + 9\sin^2 t}\cdot (-2\sin t + 3i\cos t) dt\\ = \int_{0}^{2\pi} \frac{5\sin t\cos t + 6i}{4\cos^2 t+ 9\sin^2 t}dt = 2\pi i.$$
I'm not sure where to go from here.

Robert Z
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1 Answers1

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First we rearrange the integral: $$I:=\int_{0}^{2\pi}\frac{1}{4\cos^2 t + 9\sin^2 t} dt=\int_{0}^{2\pi}\frac{1}{4 + 5\sin ^2 t} dt=\int_{0}^{2\pi}\frac{2}{13 - 5\cos 2t} dt\\=\int_{0}^{2\pi}\frac{2}{13 - 5\cos s} ds$$ where we used the fact that $2\sin^2t=1-\cos(2t)$. Now we use the substitution $z=e^{is}$: $$I:=\int_{|z|=1}\frac{2}{13 - 5\frac{z+1/z}{2}}\cdot \frac{dz}{iz} =\frac{4}{i}\int_{|z|=1}\frac{dz}{(1-5z)(z-5)}\\ =\frac{8}{\pi}\,\mbox{Res}\left(\frac{1}{(1-5z)(z-5)},\frac{1}{5}\right)=\frac{\pi}{3}.$$

P.S. As noted by MyGlasses, you are almost done: $$6\int_{0}^{2\pi} \frac{dt}{4\cos^2 t+ 9\sin^2 t}=\mbox{Im}\left(\int_{0}^{2\pi} \frac{5\sin t\cos t + 6i}{4\cos^2 t+ 9\sin^2 t}dt\right) = 2\pi$$ and again we find that $I=\frac{\pi}{3}$.

Robert Z
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