Question

Evaluate ${\int }^{}\frac{\sqrt{1+\cos 2x}}{2}dx$

Answer 1

We have,

$I=\int \frac{\sqrt{1+\cos 2x}}{2}dx$

We know that

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

Therefore,

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

$I=\int \frac{\sqrt{2{\cos }^{2}x}}{2}dx$

$I=\int \frac{\sqrt{2}\cos x}{2}dx$

$I=\frac{1}{\sqrt{2}}\int \cos xdx$

$I=\frac{1}{\sqrt{2}}\sin x+C$

Hence, this is the answer.