Calculate the integral $\int_{0}^{2\pi}\frac{1}{a^{2}\cos^2t+b^{2}\sin^{2}t}dt$, by deformation theorem.

I want to prove:

$$\int_{0}^{2\pi}\frac{1}{a^{2}\cos^2t+b^{2}\sin^{2}t}dt=\frac{2\pi}{ab}$$

by the deformation theorem of complex variable.

Then I consider a parameterization $\gamma:[0,2\pi]\rightarrow A$, traveled in the opposite direction of the clock hand, of the ellipse (in $\mathbb{C}$):

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

I thought of searching for a function $f:A\rightarrow \mathbb{C}$ (analytic in $A$) and a curve $\lambda:[0,2\pi]\rightarrow A$ (homotopic to $\gamma$) and use the deformation theorem:

$$\int_{\gamma}f = \int_{\lambda}f$$

But I can not find a function $f(z)$ that meets what I need, can you help me to find the f?

P.D: Necessarily I have to use the theorem of deformation

Let the ellipse $\Gamma$ $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

be parameterized by $$ z = a\cos\theta + ib\sin\theta$$ with $\theta\in[0,2\pi].$

Consider the integral $$ \int_\Gamma \frac{1}{z} dz.$$

This is equal to $$ \int_0^{2\pi} \frac{1}{a\cos\theta + ib\sin\theta} (-a\sin\theta + ib\cos\theta) \; d\theta\\ = \int_0^{2\pi} \frac{a\cos\theta-ib\sin\theta}{a^2\cos^2\theta + b^2\sin^2\theta} (-a\sin\theta + ib\cos\theta) \; d\theta.$$ which is $$\int_0^{2\pi} \frac{(b^2-a^2)\sin\theta\cos\theta + iab(\cos^2\theta+\sin^2\theta)} {a^2\cos^2\theta + b^2\sin^2\theta} \; d\theta \\=\int_0^{2\pi} \frac{(b^2-a^2)\sin\theta\cos\theta + iab} {a^2\cos^2\theta + b^2\sin^2\theta} \; d\theta.$$

This implies that $$ \frac{1}{ab} \Im \int_\Gamma \frac{1}{z} dz = \int_0^{2\pi} \frac{1} {a^2\cos^2\theta + b^2\sin^2\theta} \; d\theta.$$

Now apply the deformation theorem, turning the ellipse into a circle round the origin of radius one, getting $$ \frac{1}{ab} \Im \int_\Gamma \frac{1}{z} dz = \frac{1}{ab} \Im \int_{|z|=1} \frac{1}{z} dz = \frac{1}{ab} \Im (2\pi i) = \frac{2\pi}{ab}.$$

Verification. We can verify this result using the Cauchy residue theorem directly. Put $z=e^{i\theta}$ so that $dz = i e^{i\theta} \; d\theta$ and $d\theta = 1/(iz) \; dz$ to get $$\int_0^{2\pi} \frac{1} {a^2\cos^2\theta + b^2\sin^2\theta} \; d\theta = \int_{|z|=1} \frac{4}{a^2(z+1/z)^2 - b^2(z-1/z)^2} \frac{1}{iz} \; dz.$$ This gives $$\frac{1}{i}\int_{|z|=1} \frac{4z}{a^2(z^2+1)^2 - b^2(z^2-1)^2} \; dz$$ This has the following four simple poles: $$\rho_{0,1} = \pm i\sqrt{\frac{a-b}{a+b}} \quad\text{and}\quad \rho_{2,3} = \pm i\sqrt{\frac{a+b}{a-b}}.$$ Now suppose that $a>b>0$ (the other cases are treated similarly). This leaves only $\rho_{0,1}$ inside the unit circle.

The residues are easy to calculate: $$\mathrm{Res}\left(\frac{4z}{a^2(z^2+1)^2 - b^2(z^2-1)^2}; z=\rho_{0,1} \right) = \left.\frac{4z}{2a^2(z^2+1)2z - 2b^2(z^2-1)2z}\right|_{z=\rho_{0,1}} \\ = \left.\frac{2}{2a^2(z^2+1) - 2b^2(z^2-1)}\right|_{z=\rho_{0,1}} = \left.\frac{1}{(a^2-b^2)z^2 + (a^2+b^2)}\right|_{z=\rho_{0,1}}.$$ Note that $$\rho_{0,1}^2 = - \frac{a-b}{a+b}$$ giving for the residues $$ \frac{1}{-(a^2-b^2)(a-b)/(a+b) + (a^2+b^2)} = \frac{1}{-(a-b)^2 + (a^2+b^2)} = \frac{1}{2ab}.$$ Therefore the value of the integral is $$2\pi i \times \frac{1}{i} \times \left(\frac{1}{2ab} + \frac{1}{2ab}\right)= \frac{2\pi}{ab},$$ QED.

Pick $\lambda$ to be the unit circle transversed one time counterclockwise:

$\displaystyle \lambda (t) = \cos t + i \sin t$,

and $f(z) = 1/z$.

$f$ is homolomorphic on the open set $\mathbb{C} \setminus \{ 0 \}$, in which both $\gamma$ and $\lambda$ are homotopic. We have by definition:

$\displaystyle \int_{\gamma} f = \int_{0}^{2\pi} f(\lambda(t)) \dot{\lambda}(t) dt$

In details, we have:

$\displaystyle f(\lambda(t)) = \frac{1}{\displaystyle a \cos t + i b \sin t}$ and $\displaystyle \dot{\lambda} (t) = - a \sin t + i b \cos t$

Now, putting it all together in the integral and multiplying by the denominator's complex conjugate, you may check that:

$\displaystyle \int_{\gamma} f = \int_{0}^{2\pi} (\ldots) + i \int_{0}^{2\pi} \frac{ab}{\displaystyle a^{2} \cos^{2} t + b^{2} \sin^{2} t} dt$

On the other hand, you know that $\int_{\lambda} f = 2 \pi i$. Since by Cauchy's theorem homotopic version both integrals must coincide, you get the desired result by equaling their imaginary parts.