Integrate $\int_0^{\pi/2}\frac{\cos^2x}{a\cos^2x + b\sin^2x}\,dx$

Question

I don't know how to deal with this integral $$I=\displaystyle\int_0^{\pi/2}\frac{\cos^2x}{a\cos^2x + b\sin^2x}\,dx$$ I reached the step

$$I =\displaystyle\ \int_0^{\pi/2}\frac{1}{a + b\tan^2x}dx$$

Now what should I do? Please help.

Answer 1

Now substitute

$\displaystyle\ u=\tan x \implies x= tan^{-1} u \implies dx=\frac{1}{1+u^2} du$

$\displaystyle\ =\int_0^{\infty}\frac{1}{(1+u^2)(a + bu^2)}du$

Partial Fractions

$\displaystyle\ =\int_0^{\infty}\frac{1}{(1+u^2)(a + bu^2)}du= \frac{1}{a-b} \left(\int_{0}^{\infty} \frac{du}{1+u^{2}} - b \int_{0}^{\infty} \frac{du}{a+bu^{2}} \right)$

Evaluating

$\displaystyle\ =\frac{1}{a-b} \left(\int_{0}^{\infty} \frac{du}{1+u^{2}} - b \int_{0}^{\infty} \frac{du}{a+bu^{2}} \right)$

$\displaystyle\ =\frac{1}{a-b} \left([\tan x]_0^{\infty} - b \left(\frac{\tan^{-1}\left[\frac{\sqrt{b}x}{\sqrt{a}}\right)}{\sqrt{ab}} \right]_0^{\infty}\right)$

Put the Values

Answer 2

put $\displaystyle I = \int^{\frac{\pi}{2}}\frac{\cos^2 x}{a\cos^2 x+b\sin^2 x}dx$ and $\displaystyle J = \int^{\frac{\pi}{2}}_{0}\frac{\sin^2 x}{a\cos^2 x+b\sin^2 x}dx$

$\displaystyle aI+bJ = \frac{\pi}{2}$

Answer 3

Continue with \begin{align} \int_0^{\pi/2}\frac{1}{a + b\tan^2x}dx = &\ \frac1{a-b}\int_0^{\pi/2}\bigg(1- \frac{b\sec^2x}{a + b\tan^2x}\bigg)dx\\ =& \ \frac1{a-b}\bigg( \frac\pi2- \frac\pi2 \sqrt{\frac ba}\bigg) =\frac\pi{2(a+\sqrt{ab})} \end{align}