Question

Evaluate the value of : ${\int }_0^{\frac{\pi }{2}}\frac{1}{1+\cos x}dx$

Answer 1

Given, ${\int }_0^{\frac{\pi }{2}}\frac{1}{1+\cos x}dx$

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

$={\int }_0^{\frac{\pi }{2}}\frac{1}{2{\cos }^2\frac{x}{2}}dx$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}{\sec }^2\left(\frac{x}{2}\right)dx$

Let $\frac{x}{2}=u$

$\Rightarrow \frac{1}{2}dx=du$

where $x>0,u>0$ and $x>\frac{\pi }{2},u>\frac{\pi }{4}]$

$\therefore =\frac{1}{2}{\int }_0^{\frac{\pi }{2}}{\sec }^2\left(\frac{x}{2}\right)dx$

$={\int }_0^{\frac{\pi }{4}}{\sec }^{2}udu$

$=[\tan u{]}_0^{\frac{\pi }{4}}$

$=\tan \frac{\pi }{4}-\tan 0$

$=1$