Question

Evaluate the definite integral: $\int_{0}^{\frac{\pi}{2}}\frac{cos^{2}xdx}{cos^{2}x+4sin^{2}x}$

Answer 1

$Let\space I=\int_{0}^{\frac{\pi}{2}}\frac{cos^{2}xdx}{cos^{2}x+4sin^{2}x}$

$⇒I=\int_{0}^{\frac{\pi}{2}}\frac{cos^{2}x}{cos^{2}x+4(1-cos^{2}x)}dx$

$⇒I=∫^{\frac{π}{2}}_0\frac{cos^2x}{cos^2x+4-4cos^2x}dx$

$⇒I=\frac{-1}{3}∫^{\frac{π}{2}}_0 \frac{4-3cos^2x-4}{4-3cos^2x}dx$

$⇒I=\frac{-1}{3}∫^{\frac{π}{2}}_0 \frac{4-3cos^2x}{4-3cos^2x}dx+\frac{1}{3}∫^{\frac{π}{2}}_0\frac{4}{4-3cos^2x}dx$

$⇒I=\frac{-1}{3}∫^{\frac{π}{2}}_0 1dx+\frac{1}{3}∫^{\frac{π}{2}}_0 \frac{4sec^2x}{4sec^2x-3}dx$

$⇒I=\frac{-1}{3}[x]^{\frac{π}{2}}_0+\frac{1}{3}∫^{\frac{π}{2}}_0 \frac{4sec^2x}{4(1+tan^2x)-3}dx$

$⇒I=-\frac{π}{6}+\frac{2}{3}∫^{\frac{π}{2}}_0 \frac{2sec^2x}{1+4tan^2x}dx...(1)$

Consider,$∫^\frac{π}{2}_0 \frac{2sec^2x}{1+4tan^2x}dx$

Let $2tanx=t⇒2sec^2xdx=dt$

When $x=0,t=0$ and when $x=\frac{π}{2},t=∞$

$⇒∫^{\frac{π}{2}}_0 \frac{2sec^2x}{1+4tan^2x}dx=∫^∞_0\frac{dt}{1+t^2}$

$=[tan^{-1}t]^∞_0$

$=[tan^{-1}(∞)-tan^{-1}(0)]$
$=\frac{π}{2}$

Therefore,from(1),we obtain

$I=-\frac{π}{6}+\frac{2}{3}[\frac{π}{2}]=\frac{π}{3}-\frac{π}{6}=\frac{π}{6}$