Question

Evaluate the definite integral:

${\displaystyle {\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2}xdx}{{\cos }^{2}x+4{\sin }^{2}x}}$

Answer 1

Let ${\displaystyle I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2}x}{{\cos }^{2}x+4{\sin }^{2}x}}$dx

${\displaystyle {\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2}x}{{\cos }^{2}x+4(1-{\cos }^{2}x)}}$dx

${\displaystyle ={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2}x}{4-3{\cos }^{2}x}}$dx

${\displaystyle =-\frac{1}{3}{\int }_0^{\pi /2} \frac{4-3{\cos }^{2}x-4}{4-3{\cos }^{2}x}}$dx

${\displaystyle =-\frac{1}{3}{\int }_0^{\frac{\pi }{2}}(1-\frac{4}{4-3{\cos }^{2}x})}$dx

${\displaystyle =-\frac{1}{3}{\int }_0^{\frac{\pi }{2}}1\cdot dx+\frac{4}{3}{\int }_0^{\frac{\pi }{2}}\frac{dx}{4-3{\cos }^{2}x}}$

${\displaystyle =-\frac{1}{3}\left(\frac{\pi }{2}\right)+\frac{4}{3}{\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{2}x}{4{\sec }^{2}x-3}}$dx

${\displaystyle =-\frac{\pi }{6}+\frac{4}{3}{\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{2}x}{4(1+{\tan }^{2}x-3)}}$dx

⇒ Put tan x = t

sec² x dx = dt

When x = 0, t = 0 and when x = ${\displaystyle \frac{\pi }{2},t=\infty }$

I = ${\displaystyle -\frac{\pi }{6}+\frac{4}{3}{\int }_0^{\infty }\frac{dt}{4(1+t^2)-3}}$

${\displaystyle =\frac{\pi }{6}+\frac{4}{3}{\int }_0^{\infty }\frac{dt}{4t^2+1}}$

${\displaystyle =-\frac{\pi }{6}+\frac{4}{3}\cdot \frac{1}{4}{\int }_0^{\infty }\frac{dt}{t^2+\frac{1}{4}}}$

${\displaystyle =-\frac{\pi }{6}+\frac{1}{3}\cdot 2{\left[{\tan }^{-1} \frac{t}{1/2}\right]}_0^{\infty }}$

${\displaystyle =-\frac{\pi }{6}+\frac{2}{3}\cdot {\left[{\tan }^{-1}2t\right]}_0^{\infty }}$

${\displaystyle =-\frac{\pi }{6}+\frac{2}{3}[{\tan }^{-1}\infty -{\tan }^{-1}0]}$

${\displaystyle =-\frac{\pi }{6}+\frac{2}{3}\cdot [\frac{\pi }{2}-0]}$

${\displaystyle =-\frac{\pi }{6}+\frac{\pi }{3}}$

${\displaystyle =\frac{\pi }{6}}$