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.
On
Hint First compute $$ I=\int_0^{2\pi} \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt=\int_0^{2\pi} \frac{\cos^2(t)-\sin^2(t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt=\int_0^{2\pi}R(\cos(t), \sin(t)) dt $$ Where $R$ is the rational function given by $$R(x,y)=\frac{x^2-y^2}{a^2y^2+b^2x^2}$$ How to do this? Put $z=e^{it}=\cos(t)+i\sin(t)$, thus $$ \cos(t)=Re(z)=\frac{z+z^{-1}}{2}, \ \\ \sin(t)=Im(z)=\frac{z-z^{-1}}{2i}, \\ dz=ie^{it} dt=iz \ dt \Longrightarrow dt=\frac{1}{iz}dz $$ Then $I$ can be seen as a contour integral, solve it by using residues $$ I=\int_0^{2\pi}R(\cos(t), \sin(t)) dt= \int_{|z|=1} R\left(\frac{z+1/z}{2}, \frac{z-1/z}{2i} \right)\frac{1}{iz}dz $$ Finally: Note that $$ \int_0^{\pi} \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt=\frac{I}{2} $$
On
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 $$ \int_0^\pi \frac{\cos 2t}{b^2\cos^2t+a^2\sin^2t}\,dt=\frac{(a-b)\pi}{ab(a+b)}. $$
One trick to facilitate analysis is to write
$$\sin^2x=\frac{1-\cos 2x}{2}$$
and
$$\cos^2x=\frac{1+\cos 2x}{2}$$
Thus,
$$a^2\sin^2(t)+b^2\cos^2(t)=\frac{b^2+a^2}{2}+\frac{b^2-a^2}{2}\cos 2t$$
For the integral of interest, we can write
$$\begin{align} I_1&=\int_0^{\pi} \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt\\\\ &=\frac{1}{2}\int_0^{2\pi}\frac{\cos t}{\frac{b^2+a^2}{2}+\frac{b^2-a^2}{2}\cos t}dt\\\\ &=\frac{1}{(b^2-a^2)}\int_0^{2\pi}\frac{\cos t}{\frac{b^2+a^2}{b^2-a^2}+\cos t}dt\\\\ &=\frac{2\pi}{(b^2-a^2)}-\frac{b^2+a^2}{(b^2-a^2)^2}\int_0^{2\pi}\frac{1}{\frac{b^2+a^2}{b^2-a^2}+\cos t}dt \end{align}$$
Can you finish now?
SPOLIER ALERT SCROLL OVER SHADED AREA TO SEE ANSWER
NOTE:
For $I_2=\int_0^{\pi} \frac{\sin(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt$, enforce the substitution $t= \pi -x$. Then, we see
$$\begin{align} I_2&=\int_0^{\pi} \frac{\sin(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt\\\\ &=\int_{\pi}^{0} \frac{\sin 2(\pi-x)}{a^2\sin^2(\pi-x)+b^2\cos^2(\pi-x)}(-1)dx\\\\ &=\int_{0}^{\pi} \frac{-\sin (2x)}{a^2\sin^2(x)+b^2\cos^2(x)}dx\\\\ &=-I_2 \end{align}$$
Thus, we have $I_2=-I_2$ which, of course, implies $I_2=0$!