figure out $\int_{-\pi}^{\pi}\frac{ab}{a^2\cos^2 t+b^2\sin^2 t}\,dt$

Question

$$\int_{-\pi}^{\pi}\frac{ab}{a^2\cos^2 t+b^2\sin^2 t}\,dt$$ We are learning complex analysis... And an exercise is to verify $\int_{\gamma}z^{-1}dz=2\pi i$ where $\gamma(t)=a\cos t+i b\sin t$. I do some algebra and is stuck on the integral above.. I know we can use residue theorem to easily get the answer. I was wondering is any way in real analysis to figure this out?

Answer 1

In THIS ANSWER, I used contour integration along with notes on other approaches to find the value of the integral of interest.

Here, the OP requested a way forward using real analysis only. To that end, we proceed with a ...

HINT:

$$\begin{align} \int_{-\pi}^\pi\frac{ab}{a^2\cos^2 t+b^2\sin^2 t}\,dt&=4\int_{0}^{\pi/2}\frac{(b/a)\,\sec^2 t}{1+(b/a)^2\tan^2 t}\,dt \end{align}$$

Now substitute $u=(b/a)\tan t$, with $du=(b/a)\sec^2 t\,dt$.