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How I can find the anti-derivative? $$\int\frac{\sin x}{1+x\cos x}dx$$

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  • $\begingroup$ Looks ugly. $\endgroup$ – user61527 Apr 16 '14 at 6:02
  • $\begingroup$ Have you tried using the simplification of expressions option? $\endgroup$ – mahmoud afefey Apr 16 '14 at 6:16
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    $\begingroup$ No. What is that? $\endgroup$ – Hamid Shafie Asl Apr 16 '14 at 6:16
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    $\begingroup$ And where do you arrive ? $\endgroup$ – Claude Leibovici Apr 16 '14 at 7:11
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    $\begingroup$ If no bounds, I am skeptical concerning the existence of a closed form solution. $\endgroup$ – Claude Leibovici Apr 16 '14 at 7:14
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By using the formula http://upload.wikimedia.org/wikipedia/en/math/7/2/a/72a1058ad2087aec467af24bddcf9479.png, we have $\int\dfrac{\sin x}{1+x\cos x}dx=x\sum\limits_{n=1}^\infty\sum\limits_{m=1}^{2^n-1}\dfrac{(-1)^{m+1}\sin\dfrac{mx}{2^n}}{2^n+mx\cos\dfrac{mx}{2^n}}+C$

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Sorry for “cheating”, but it is as it seems: Wolfram|Alpha states it is not solvable “in terms of standard mathematical functions” (which should then be true).

http://www.wolframalpha.com/input/?i=%E2%88%ABsinx%2F%281%2Bxcosx%29dx

Are you sure this is the correct integral?

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    $\begingroup$ There are many integrals that Wolfram Alpha can't solve, but which have solutions. That being said this is probably not one of them... $\endgroup$ – Winther Jul 1 '14 at 6:20
  • $\begingroup$ Strange. I am used to Wolfram Alpha having solved way more complex integrals… seems I am going to grab a piece of paper now. $\endgroup$ – Lukas Juhrich Jul 1 '14 at 6:22

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