TITLE: Integral of $\sin(\sqrt{x})$
QUESTION [6 upvotes]: I need help finding the integral of $\sin(\sqrt{x})dx$. I have the answer here but would like to know how to get there.

REPLY [9 votes]: Use the substitution $t=\sqrt{x}$, and then use integration by parts.  From the substitution, you should get:
$$\int\sin\sqrt{x}dx=2\int t\sin (t) dt$$
Then, use integration by parts.  Let $u=t$ and $dv=\sin t$.  Then $du=dt$ and $v=-\cos t$, and we have:
$$\int t\sin (t) dt=-t\cos t +\int \cos (t) dt=\sin t -t\cos t=\sin\sqrt{x}-\sqrt{x}\cos\sqrt{x}$$
and we have the solution:
$$\int\sin\sqrt{x}dx=2(\sin\sqrt{x}-\sqrt{x}\cos\sqrt{x})$$
I hope this clears up the matter for you!