Evaluate:

Let $I=\int \sin x \sqrt{(1+\cos 2 x)} d x$

$=\int \sin x \sqrt{(1+\cos 2 x)} d x$

$=\int \sin x \sqrt{2 \cos ^{2} x} d x$

$=\int \sin x \sqrt{2} \cos x d x$

$=\sqrt{2} \int \sin x \cos x d x$

Now, Multiply and Divide by 2 we get,

$=\frac{\sqrt{2}}{2} \int 2 \sin x \cos x d x$

$=\frac{\sqrt{2}}{2} \int \sin 2 x d x$

$=\frac{\sqrt{2}}{2} \frac{-\cos 2 x}{2}$

Hence, $I=-\frac{1}{2 \sqrt{2}} \cos 2 \mathrm{x}+\mathrm{C}$