Question

Evaluate:
${\int }_0^{\pi /2}\frac{dx}{(4{\sin }^{2}x+5{\cos }^{2}x)}$

Answer 1

Consider the given integral.

$I={\int }_0^{\pi /2}\frac{dx}{(4{\sin }^{2}x+5{\cos }^{2}x)}$

$I={\int }_0^{\pi /2}\frac{dx}{(4(1-{\cos }^{2}x)+5{\cos }^{2}x)}$

$I={\int }_0^{\pi /2}\frac{dx}{(4-4{\cos }^{2}x+5{\cos }^{2}x)}$

$I={\int }_0^{\pi /2}\frac{dx}{(4+{\cos }^{2}x)}$

$I={\int }_0^{\pi /2}\frac{{\sec }^{2}xdx}{(4{\sec }^{2}x+1)}$

$I={\int }_0^{\pi /2}\frac{{\sec }^{2}xdx}{(4(1+{\tan }^{2}x)+1)}$

$I={\int }_0^{\pi /2}\frac{{\sec }^{2}xdx}{(4+4{\tan }^{2}x+1)}$

$I=\frac{1}{4}{\int }_0^{\pi /2}\frac{{\sec }^{2}xdx}{(\frac{5}{4}+{\tan }^{2}x)}$

Let $t=\tan x$

$dt={\sec }^{2}xdx$

Therefore,

$I=\frac{1}{4}{\int }_0^{\infty }\frac{1}{{\left(\frac{\sqrt{5}}{2}\right)}^2+t^2}dt$

$I=\frac{1}{4}{\left[\frac{1}{\frac{\sqrt{5}}{2}}{\tan }^{-1}\left(\frac{t}{\frac{\sqrt{5}}{2}}\right)\right]}_0^{\infty }$

$I={\left[\frac{1}{2\sqrt{5}}{\tan }^{-1}\left(\frac{2t}{\sqrt{5}}\right)\right]}_0^{\infty }$

$I=[\frac{1}{2\sqrt{5}}{\tan }^{-1}\left(\infty \right)-\frac{1}{2\sqrt{5}}{\tan }^{-1}\left(0\right)]$

$I=\frac{1}{2\sqrt{5}}{\tan }^{-1}\left(\tan \frac{\pi }{2}\right)-\frac{1}{2\sqrt{5}}{\tan }^{-1}\left(\tan 0\right)$

$I=\frac{\pi }{4\sqrt{5}}-0$

$I=\frac{\pi }{4\sqrt{5}}$

Hence, this is the answer.