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This question already has an answer here:

How do I evaluate:

$$\int_{0}^{\pi} \sin (\sin x) \ dx$$

I have seen a similar question here but can't find it.

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marked as duplicate by M.H, vadim123, Julian Kuelshammer, Davide Giraudo, azimut May 10 '13 at 13:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ math.stackexchange.com/questions/287395/… $\endgroup$ – Dimitris May 10 '13 at 13:07
  • $\begingroup$ It does not appear that the area under this curve can be expressed with simple functions. Are you sure the upper bound isn't $2\pi$? $\endgroup$ – apnorton May 10 '13 at 13:10
  • $\begingroup$ I'm sure it is $\pi$ but what will happen if we change it to $2\pi$? $\endgroup$ – please delete me May 10 '13 at 13:12
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    $\begingroup$ @AlexanderJones If you integrate from $0$ to $2\pi$, you're integrating an odd periodic function over its period--the answer is $0$. (Which makes the problem a lot easier...) $\endgroup$ – apnorton May 10 '13 at 13:14
  • $\begingroup$ This is not a duplicate of that question. This asks for a definite integral, whereas that asks for an indefinite integral. A definite answer to that question would provide an answer to this question, but that question only seems to have suggestions. $\endgroup$ – robjohn May 10 '13 at 15:57
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This is not a closed form (I don't know if one exists in elementary functions), but this series converges pretty fast: $$ \begin{align} \int_0^\pi\sin(\sin(x))\,\mathrm{d}x &=2\int_0^1\frac{\sin(u)}{\sqrt{1-u^2}}\,\mathrm{d}u\\ &=2\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\int_0^1\frac{u^{2k+1}}{\sqrt{1-u^2}}\,\mathrm{d}u\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\int_0^1\frac{v^k}{\sqrt{1-v}}\,\mathrm{d}v\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\frac{\Gamma(k+1)\Gamma(1/2)}{\Gamma(k+3/2)}\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\frac{k!\sqrt\pi}{(k+1/2)k!\binom{2k}{k}\frac{\sqrt\pi}{4^k}}\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\frac{2^{2k+1}}{(2k+1)\binom{2k}{k}} \end{align} $$

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When you substitute $u=\sin{x}$, you get

$$2 \int_0^1 du \frac{\sin{u}}{\sqrt{1-u^2}}$$

The integral is related to a Struve function:

$$\mathbf{H}_0(z) = \frac{2}{\pi} \int_0^1 du \frac{\sin{z u}}{\sqrt{1-u^2}}$$

Then the integral is equal to $\pi \mathbf{H}_0(1)$.

See also my solution to a similar question.

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  • $\begingroup$ When I enter my series into Mathematica 8, it recognizes it as π StruveH[0,1] $\endgroup$ – robjohn May 10 '13 at 14:25
  • $\begingroup$ @robjohn: That is correct and equivalent to what I have written. $\endgroup$ – Ron Gordon May 10 '13 at 14:27
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    $\begingroup$ Yeah, I was just noting that Mathematica 8 recognized the function you cited from the series I cited. :-) $\endgroup$ – robjohn May 10 '13 at 14:28

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