Question

Show that ${\int }_0^{\pi /2}\frac{{\cos }^2\theta d\theta }{{\cos }^2\theta +4{\sin }^2\theta }=\frac{\pi }{2}$.

Answer 1

Let $I={\int }_0^{\pi /2}\frac{{\cos }^2\theta d\theta }{{\cos }^2\theta +4{\sin }^2\theta }$
Multiplying numerator and denominator by ${\sec }^4\theta$, we get
$I={\int }_0^{\pi /2}\frac{{\sec }^2\theta d\theta }{{\sec }^2\theta +4{\tan }^2\theta {\sec }^2\theta }d\theta$

$={\int }_0^{\pi /2}\frac{{\sec }^2\theta d\theta }{4{\tan }^4\theta +5{\tan }^2\theta +1}d\theta$
Put $t=\tan \theta \Rightarrow dt={\sec }^2\theta d\theta$

When $\theta =0\Rightarrow t=$

When $\theta =\frac{\pi }{2}\Rightarrow t=\infty$

Therefore
$I={\int }_0^{\infty }\frac{1}{4t^4+5t^2+1}dt$

$I={\int }_0^{\infty }\frac{1}{(4t^2+1)(t^2+1)}dt$ .....(1)

Resolving $\frac{1}{(4t^2+1)(t^2+1)}$ into partial fractions

Let $t^2=u$

$\frac{1}{(4u+1)(u+1)}=\frac{A}{4u+1}+\frac{B}{u+1}$ ....(2)

$\Rightarrow 1=A(u+1)+B(4u+1)$

When $u=-1\Rightarrow B=-\frac{1}{3}$

When $u=0\Rightarrow A=\frac{4}{3}$

Put these values in (2)

$\frac{1}{(4u+1)(u+1)}=\frac{4}{3(4u+1)}-\frac{1}{3(u+1)}$

$\frac{1}{(4t^2+1)(t^2+1)}=\frac{4}{3(4t^2+1)}-\frac{1}{3(t^2+1)}$

Using this in eqn (1), we get

$I={\int }_0^{\infty }(\frac{4}{3(4t^2+1)}-\frac{1}{3(t^2+1)})dt$
$={\left[\frac{2}{3}{\tan }^{-1}2t\right]}_0^{\infty }+{\left[\frac{1}{3}{\tan }^{-1}t\right]}_0^{\infty }=\frac{\pi }{2}$