integral $\int_0^\pi \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt$ I want to compute this integral
$$\int_0^\pi \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt$$
where $0<b \leq a$.
I have this results
$$I_1=-\frac{ab}{2\pi}\int_0^\pi \frac{\cos(2t)}{a^2sin^2(t)+b^2cos^2(t)}dt=\frac{a}{a+b}-\frac{1}{2}$$
But I don't know how to prove this equality.
Which can help me,
Thanks for all.

Although there is already an accepted answer, I put this answer here, since someone might have use of the "tool" to work with the expression like this before integrating.
The case $a=b$ is trivial, so we assume $a>b$. We write the numerator $\cos 2t$ in the following way:
$$
\cos 2t=\cos^2t-\sin^2t=\alpha\bigl(b^2\cos^2t+a^2\sin^2t\bigr)+(1-\alpha b^2)\cos^2t-(1+\alpha a^2)\sin^2t,
$$
and we'll soon see how to choose $\alpha$ in a good way, where good means that we should get something that we can integrate. Let us first use the trig-one on the $\cos^2 t$ part, to get
$$
\cos 2t = \alpha\bigl(b^2\cos^2t+a^2\sin^2t\bigr)+(1-\alpha b^2)-\bigl[(1-\alpha b^2)+(1+\alpha a^2)\bigr]\sin^2 t.
$$
Now, we choose $\alpha$ so that the constant in brackets is zero, i.e.
$$
(1-\alpha b^2)+(1+\alpha a^2)=0 \iff \alpha=\frac{2}{b^2-a^2}.
$$
This gives
$$
\cos 2t = \frac{2}{b^2-a^2}(b^2\cos^2t+a^2\sin^2t)-\frac{a^2+b^2}{b^2-a^2}.
$$
Now comes the fun part. We can write the integrand
$$
\begin{aligned}
\frac{\cos 2t}{b^2\cos^2t+a^2\sin^2t}&=\frac{2}{b^2-a^2}-\frac{a^2+b^2}{b^2-a^2}\frac{1}{b^2\cos^2t+a^2\sin^2t}\\
&=\frac{2}{b^2-a^2}-\frac{a^2+b^2}{b^2-a^2}\frac{1}{b^2\cot^2t+a^2}\frac{1}{\sin^2t}.
\end{aligned}
$$
This is easy to integrate,
$$
\int \frac{\cos 2t}{b^2\cos^2t+a^2\sin^2t}\,dt = \frac{2t}{b^2-a^2}-\frac{a^2+b^2}{b^2-a^2} \frac{1}{ab}\text{arccot}\,\Bigl(\frac{b\cot t}{a}\Bigr).
$$
I'm sure you can insert the limits and get the result from this. I get
$$
\int_0^\pi \frac{\cos 2t}{b^2\cos^2t+a^2\sin^2t}\,dt=\frac{(a-b)\pi}{ab(a+b)}.
$$
