Question

Solve;
${\int }_0^{\frac{\pi }{2}}\frac{1}{1+{\cos }^{2}x}dx$

Answer 1

${\int }_0^{\frac{\pi }{2}}\frac{dx}{1+{\cos }^{2}x}={\int }_0^{\frac{\pi }{2}}\frac{dx}{{cosec}^{2}x}={\int }_0^{\frac{\pi }{2}}{\sin }^{2}xdx={\int }_0^{\frac{\pi }{2}}\frac{1-\cos 2x}{2}dx=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}(1-\cos 2x)dx=\frac{1}{2}{[x-\frac{\sin x}{2}]}^{\frac{\pi }{2}}=\frac{1}{2}[\frac{\pi }{2}-\frac{\sin 2\times \frac{\pi }{2}}{2}]=\frac{1}{2}[\frac{\pi }{2}-0]=\frac{\pi }{4}$