Question

Evaluate ${\displaystyle {\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2}x}{1+\sin x\cos x}dx}$

Answer 1

${\displaystyle I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2}x}{1+\sin x\cos x}dx}$   ....(1)

Using ${\displaystyle {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx}$

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

${\displaystyle ={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}x}{1+\cos x.\sin x}dx}$  .....(2)

Adding eq. (1) & (2)

${\displaystyle 2I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2}x+{\sin }^{2}x}{1+\sin x\cos x}dx}$

${\displaystyle ={\int }_0^{\frac{\pi }{2}}\frac{1}{1+\sin x\cos x}dx}$`

${\displaystyle ={\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{2}x}{{\sec }^{2}x+\tan x}dx}$

${\displaystyle 2I={\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{2}xdx}{1+{\tan }^{2}x+\tan x}}$

Put ${\displaystyle \tan x=t,{\sec }^{2}xdx=dt}$

when x = 0, t = 0

when ${\displaystyle s=\frac{\pi }{2},t=\infty }$

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

${\displaystyle ={\int }_0^{\infty } \frac{dt}{{(t+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}}$

${\displaystyle =\frac{1}{\frac{\sqrt{3}}{2}}{\left[{\tan }^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)\right]}_0^{\infty }}$

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

${\displaystyle 2I=\frac{2}{\sqrt{3}}[\frac{\pi }{2}-\frac{\pi }{6}]}$

${\displaystyle I=\frac{1}{\sqrt{3}}\left[\frac{3\pi -\pi }{6}\right]}$

${\displaystyle =\frac{1}{\sqrt{3}}\left[\frac{2\pi }{6}\right]=\frac{\pi }{3\sqrt{3}}}$