TITLE: How to evaluate the integral of $\sqrt{\sin\sqrt x}\cos \sqrt x / ( 1+x^2)$?
QUESTION [6 upvotes]: $$
\int \frac{\sqrt{\sin\sqrt x}\cos \sqrt x}{1+x^2} dx
$$
I have tried combinations of $x=t^2$, integration by parts, $\tan\left(\dfrac u2\right)$ substitutions it got even more complicated. Is there a way to evaluate this integral with elementary techniques?

REPLY [2 votes]: Hint:
Let $t=\sqrt x$ ,
Then $x=t^2$
$dx=2t~dt$
$\therefore\int\dfrac{\sqrt{\sin\sqrt x}\cos\sqrt x}{1+x^2}dx$
$=\int\dfrac{\sqrt{\sin t}\cos t}{t^4+1}dt$
$=\int\dfrac{\sqrt{\sin t}}{t^4+1}d(\sin t)$
$=\int\dfrac{2}{3(t^4+1)}d\left(\sin^\frac{3}{2}t\right)$
$=\dfrac{2\sin^\frac{3}{2}t}{3(t^4+1)}-\int\sin^\frac{3}{2}t~d\left(\dfrac{2}{3(t^4+1)}\right)$
$=\dfrac{2\sin^\frac{3}{2}t}{3(t^4+1)}+\int\dfrac{8t^3\sin^\frac{3}{2}t}{3(t^4+1)^2}dt$