Evaluate $\int \frac{\sqrt{\sin\sqrt x}\cos \sqrt x}{1+x^2} dx$?

Question

$$ \int \frac{\sqrt{\sin\sqrt x}\cos \sqrt x}{1+x^2} dx $$

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?

Answer 1

Hint:

Let $t=\sqrt x$ ,

Then $x=t^2$

$dx=2t~dt$

$\therefore\int\dfrac{\sqrt{\sin\sqrt x}\cos\sqrt x}{1+x^2}dx$

$=\int\dfrac{\sqrt{\sin t}\cos t}{t^4+1}dt$

$=\int\dfrac{\sqrt{\sin t}}{t^4+1}d(\sin t)$

$=\int\dfrac{2}{3(t^4+1)}d\left(\sin^\frac{3}{2}t\right)$

$=\dfrac{2\sin^\frac{3}{2}t}{3(t^4+1)}-\int\sin^\frac{3}{2}t~d\left(\dfrac{2}{3(t^4+1)}\right)$

$=\dfrac{2\sin^\frac{3}{2}t}{3(t^4+1)}+\int\dfrac{8t^3\sin^\frac{3}{2}t}{3(t^4+1)^2}dt$