# Proving a trigonometric integral by integrating around an ellipse in the complex plain

Prove for $a,b > 0$

$$\int^{2\pi}_0\frac{dt}{a^2\cos^2 t +b^2\sin^2 t} = \frac{2\pi}{ab}$$

$$\textit{solution}$$

Let us integrate the following function
$$f(z) = \frac{1}{z}$$

Around the ellipse

$$\oint_{\gamma}f(z)\,dz =2\pi i\,\mathrm{Res}(f,0)$$

The parametrization of the ellipse $\gamma(t) = a\cos(t)+ib\sin(t)$

$$\oint_{\gamma}f(z)\,dz=\int^{2\pi}_0 \frac{-a\sin t+ib\cos t}{a\cos t+ib\sin t} \,dt$$

This simplifies to the following
$$\frac{-a\sin t+ib\cos t}{a\cos t+ib\sin t} \times \frac{a\cos t-ib\sin t}{a\cos t-ib\sin t} = \frac{(b^2-a^2)\sin t \cos t+iab}{a^2\cos^2 t +\sin^2 t}$$

Hence

$$\int^{2\pi}_0\frac{(b^2-a^2)\sin t \cos t+iab}{a^2\cos^2 t +b^2\sin^2 t}\,dt = 2\pi i \, \mathrm{Res}(f,0) = 2\pi i$$

By equating the imaginary parts

$$iab\int^{2\pi}_0\frac{dt}{a^2\cos^2 t +b^2\sin^2 t} = 2\pi i$$

Which implies the result

$$\int^{2\pi}_0\frac{dt}{a^2\cos^2 t +b^2\sin^2 t} = \frac{2\pi}{ab}$$

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