Question

Solve ${\int }_0^{\pi /4}\frac{dx}{1+\cos 2x}$

Answer 1

We have,

$I={\int }_0^{\frac{\pi }{4}}\frac{dx}{1+\cos 2x}$

Since,

$\cos 2x=2{\cos }^{2}x-1$

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

$I=\frac{1}{2}{\int }_0^{\frac{\pi }{4}}\frac{dx}{{\cos }^{2}x}$

$I=\frac{1}{2}{\int }_0^{\frac{\pi }{4}}{\sec }^{2}xdx$

$I=\frac{1}{2}{\left[\tan x\right]}_0^{\frac{\pi }{4}}$

$I=\frac{1}{2}[\tan \frac{\pi }{4}-\tan 0]$

$I=\frac{1}{2}$

$I=0.5$

Hence, this is the answer.