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

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?

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    $\begingroup$ It is pretty tough, as I can not see an integral at all.. is the answer $\pi$ or 42 ;)? seeing as someone has upvoted .. i guess people can see it!! $\endgroup$ – Chinny84 Nov 27 '14 at 12:06
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    $\begingroup$ WolframAlpha times out, so conventional methods probably won't work. There might be some clever substitution in there, though. $\endgroup$ – Arthur Nov 27 '14 at 12:07
  • $\begingroup$ I don't know the answer at all. $\endgroup$ – reco Nov 27 '14 at 12:09
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    $\begingroup$ Any motivation for this integral? Where/how did you find it? $\endgroup$ – Arian Nov 27 '14 at 12:58
  • $\begingroup$ I reduced the problem to finding an integral of sin(x)^{3/2}/(x+p) for p constant. If anybody can do this I would post how I reduced the problem. $\endgroup$ – Not Buying It Nov 27 '14 at 13:37
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Hint:

Let $t=\sqrt x$ ,

Then $x=t^2$

$dx=2t~dt$

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

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

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

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

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

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

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