A typical integral in complex analysis

Show that $$\int_{0}^{2\pi}\frac{1}{a^{2}\cos^2\theta+b^{2}\sin^{2}\theta}\; d\theta=\frac{2\pi}{ab}$$

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Method I: Cauchy Integral Formula
Let $C$ be the ellipse $\{z=x+iy\mid x=a\cos \theta,\quad y=b\sin \theta,\quad 0\leq \theta \leq 2\pi\}$.

We consider the integral $$\begin{align}\int_C \frac{1}{z}\; dz&=\int_0^{2\pi} \frac{1}{a\cos\theta+ib\sin\theta}(-a\sin\theta+ib\cos\theta)\; d\theta\\&=\int_0^{2\pi} \frac{a\cos\theta-ib\sin\theta}{a^2\cos^2\theta+b^2\sin^2\theta}(-a\sin\theta+ib\cos\theta)\; d\theta\\&=\int_0^{2\pi} \frac{(b^2-a^2)\sin\theta\cos\theta + iab}{a^2\cos^2\theta + b^2\sin^2\theta}\; d\theta.\end{align}$$ Since we know that $$\int_{C} \dfrac{1}{z}dz=2\pi i,$$ we have $$\int_0^{2\pi} \frac{(b^2-a^2)\sin\theta\cos\theta}{a^2\cos^2\theta + b^2\sin^2\theta}\; d\theta=0\\ \int_0^{2\pi} \frac{iab}{a^2\cos^2\theta + b^2\sin^2\theta}\; d\theta=2\pi i.$$ We therefore have $$\int_{0}^{2\pi}\frac{\sin\theta\cos\theta}{a^2\cos^2\theta + b^2\sin^2\theta}\; d\theta=0\\ \int_{0}^{2\pi}\frac{1}{a^{2}\cos^2\theta+b^{2}\sin^{2}\theta}\; d\theta=\frac{2\pi}{ab}.$$

Method II: Residue theorem
Put $z=e^{i\theta}$. Then $dz=ie^{i\theta}\; d\theta$ and $d\theta=1/(iz)\; dz$. $$\begin{align}\int_0^{2\pi}\frac{1}{a^2\cos^2\theta + b^2\sin^2\theta}\; d\theta&=\int_{|z|=1} \frac{4}{a^2(z+1/z)^2 - b^2(z-1/z)^2} \frac{1}{iz}\; dz\\&=\frac{1}{i}\int_{|z|=1} \frac{4z}{a^2(z^2+1)^2 - b^2(z^2-1)^2} \; dz.\end{align}$$ This has the following four simple poles: $$z= \pm i\sqrt{\frac{a-b}{a+b}}, \pm i\sqrt{\frac{a+b}{a-b}}.$$ Now suppose that $0<b<a$ (other cases are treated similarly). This leaves only $\pm i\sqrt{\dfrac{a-b}{a+b}}$ inside the unit circle.

We now find the residue at $i\sqrt{\dfrac{a-b}{a+b}}$, which happens to be the same as that at $- i\sqrt{\dfrac{a-b}{a+b}}$: $$\begin{align}\mathrm{Res}\left(\frac{4z}{a^2(z^2+1)^2 - b^2(z^2-1)^2}; z_0=i\sqrt{\frac{a-b}{a+b}} \right)&=\left.\frac{4z}{2a^2(z^2+1)2z - 2b^2(z^2-1)2z}\right|_{z=z_0}\\&= \left.\frac{1}{(a^2-b^2)z^2 + (a^2+b^2)}\right|_{z=z_0}\\ &=\frac{1}{-(a^2-b^2)(a-b)/(a+b)+(a^2+b^2)}\\&=\frac{1}{-(a-b)^2 + (a^2+b^2)}\\&=\frac{1}{2ab}.\end{align}$$ Therefore $$\int_{0}^{2\pi}\frac{1}{a^2\cos^2\theta + b^2\sin^2\theta}\; d\theta=\frac{1}{i} 2\pi i \times \left(\frac{1}{2ab} + \frac{1}{2ab}\right)= \frac{2\pi}{ab}.$$