Question:
$\int \sqrt{1+\sin x} d x$
Solution:
Let $\mathrm{I}=\int \sqrt{1+\sin x} d x$
$=\int \sqrt{\left(\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}\right)} d x$
$=\int \sqrt{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}} d x=\int\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right) d x$
$=\int \sin \frac{x}{2} d x+\int \cos \frac{x}{2} d x=-2 \cos \frac{x}{2}+2 \sin \frac{x}{2}+C$
$=2\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)+C$, where $C$ is a constant
Therefore,
$\mathrm{I}=2\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)+\mathrm{C}$