Answer
$$\int \sin \sqrt x d x=2( \sin \sqrt x - \sqrt x \cos \sqrt x )+c$$
Work Step by Step
Let $u=\sqrt x$; then $du=\frac{1}{2\sqrt x}dx$. Hence, we have
$$\int \sin \sqrt x d x
=2 \int u \sin u \ du.$$
Now we use integration by parts as follows:
$$ \int u \sin u \ du=-u \cos u +\int \cos u \ du=-u \cos u + \sin u +c.$$
Hence $$\int \sin \sqrt x d x=2( \sin \sqrt x - \sqrt x \cos \sqrt x )+c$$
