Evaluate the definite integrals$\int\limits_0^\frac{\pi}{2}\frac{\cos^2xdx}{\cos^2x+4\sin^2x}$

Toolbox:
• (i)when f(x) is an integral function is substitute on t,then $f'(x)dx=dt.Hence\;\int f(x)dx=\int t.dt$
• (ii)$\int \frac{dx}{a^2+x^2}=\frac{1}{a}\tan ^{-1}(x/a)+c$
Given $I=\int\limits_0^\frac{\pi}{2}\frac{\cos^2x}{\cos^2x+4\sin^2x}dx$

$\sin ^2 x=1-\cos ^2x$

Therefore $I=\int\limits_0^\frac{\pi}{2}\frac{\cos^2x}{\cos^2x+4(1-\cos^2x)}dx$

$=\int\limits_0^\frac{\pi}{2}\frac{\cos^2x}{\cos^2x+4-4\cos^2x}dx$

Multiply and divide 3 to the numerator and denominator

$=\frac{1}{3}\int\limits_0^\frac{\pi}{2}\frac{3cos^2x}{4-3\cos^2x}dx$

Add and subtract 4 in he numerator,

$I=\frac{1}{3}\int\limits_0^\frac{\pi}{2}\frac{4-3\cos^2x-4}{4-3\cos^2x}dx$

now seperating the terms

$I=\frac{1}{3}\int\limits_0^\frac{\pi}{2}\frac{4-3\cos^2x}{4-3\cos^2x}dx-\int\limits_0^\frac{\pi}{2}\frac{4}{4-3\cos^2x}dx$

$=\frac{1}{3}\int\limits_0^\frac{\pi}{2}dx-\int\limits_0^\frac{\pi}{2}\frac{4}{4-3\cos^2x}dx$

Consider $I_1=\int \frac{4}{4-3 \cos ^2x};$Multiply and divide each term by $\cos ^2x$

$I_1=\Large\int\frac{4/ \cos^2x}{4/\cos ^2x -\frac{3\cos^2x}{\cos^2x}}dx=\int \frac{4 \sec ^2x}{4\sec^2x-3}dx$

But $\sec^2x=1+\tan^2x$

Hence $I_1=\int \frac{4 \sec ^2x}{4(1+\tan^2x)-3}dx$

Hence $I=-\frac{1}{3}\int\limits_0^\frac{\pi}{2}dx+\frac{1}{3}\int\limits_0^\frac{\pi}{2}\frac{4 \sec ^2x}{1+4\tan^2x}dx$

Consider $I_1= \frac{1}{3}\int\limits_0^\frac{\pi}{2}\frac{4 \sec ^2x dx}{1+4\tan^2x}=\frac{2}{3}\int\limits_0^\frac{\pi}{2}\frac{2 \sec ^2x\;dx}{1+4\tan^2x}$

Put $t=2\tan x$; on differentiating w.r.t x,$dt=2\sec^2xdx$

Limits also changes ;$when x=0;t=0 \;when\; x=\pi/2;t=\infty$

on substituting $\infty$

$I_1=\frac{2}{3}\int\limits_0^\infty\frac{dt}{1+t^2}$

$I=-\frac{1}{3}\int\limits_0^{\frac{\pi}{2}}dx+\frac{2}{3}\int\limits_0^\infty\frac{dt}{1+t^2}$

On integrating we get,

$I=-\frac{1}{3}[\pi/2-0+\frac{2}{3}[\tan^{-1}(t)]_0^\infty$

on applying limits we get

$I=-\frac{1}{3}[\pi/2-0]+\frac{2}{3}[\tan^{-1}(\infty)-\tan^{-1}(0)]$

$=-\frac{1}{3}(\pi/2)+\frac{2}{3}(\pi/2)$

$(tan^{-1}(\infty))=\pi/2$

Therefore $I=-\frac{\pi}{6}+\frac{2\pi}{6}=\frac{\pi}{6}$