Question

Solve:
${\int }_0^{\pi /4}\sqrt{1+\sin 2x}dx$

Answer 1

$(1+\sin 2x)={\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x$

$=(\sin x+\cos x{)}^2$

So, $\sqrt{(1+\sin 2x)}=\sin x+\cos x$

Therefore,

Integral of $(\sin x+\cos x)=$ Integration of $\sin x+$ integration of $\cos x=-\cos x+\sin x$

Putting the limits from $0$ to $\frac{\pi }{4}$, we get,

$-[\cos \frac{\pi }{4}-\cos 0]+[\sin \frac{\pi }{4}-\sin 0]$

$=-\frac{1}{\sqrt{2}}+1+\frac{1}{\sqrt{2}}-0$

$=1$

Hence, this is the answer.