Integrate $\int_0^{2\pi}\frac{1}{4\cos^2(t)+9\sin^2(t)}\mathrm{d}t$ with complex functions or Fourier transformations.

Question

What is the value of the following integral: $$\int_0^{2\pi}\frac{1}{4\cos^2(t)+9\sin^2(t)}\mathrm{d}t$$ $\frac\pi9$ ; $\frac\pi6$ ; $\frac\pi3$ ; $\frac\pi2$ or $\frac\pi4$?

a full solution for this problem would be much appreciated

Answer 1

Integrate with Fourier series as follows

$$\int_0^{2\pi}\frac{1}{4\cos^2 t+9\sin^2 t}{d}t =\int_0^{2\pi}\frac{2}{13-5\cos 2t} {d}t\\ =\int_0^{2\pi}\left( \frac16 + \frac13\sum_{n=1}^\infty \frac{1}{5^n} \cos (2n t) \right)dt =\frac\pi3 $$

Answer 2

Here is a solution using differentiation under integral sign. Consider a more general integral, namely:

\begin{align*} I_1(\alpha,\beta) & = \int_{0}^{2\pi}\frac{dx}{\alpha \cos^2x+\beta \sin^2x} \\ & = \int_{0}^{2\pi} \frac{\sec^2x}{\alpha + \beta \tan^2x} \,\mathrm{d}x\\ & = \frac{1}{\beta} \int_0^{2\pi} \frac{1}{\left(\sqrt{\frac{\alpha}{\beta}}\right)^2 + \tan^2 (x)}\; \mathrm{d}(\tan x)\,\\ & = \frac{1}{\sqrt{\alpha \beta}} \left[ \left(\tan^{-1}\left(\sqrt{\frac{\beta}{\alpha}}\tan (x)\right)\right) \right]_0^{2\pi} = \frac{2\pi}{\sqrt{\alpha\beta}} \end{align*}