Definite trig integral How do I evaluate:
$$\int_{0}^{\pi} \sin (\sin x) \ dx$$
I have seen a similar question here but can't find it.
 A: This is not a closed form (I don't know if one exists in elementary functions), but this series converges pretty fast:
$$
\begin{align}
\int_0^\pi\sin(\sin(x))\,\mathrm{d}x
&=2\int_0^1\frac{\sin(u)}{\sqrt{1-u^2}}\,\mathrm{d}u\\
&=2\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\int_0^1\frac{u^{2k+1}}{\sqrt{1-u^2}}\,\mathrm{d}u\\
&=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\int_0^1\frac{v^k}{\sqrt{1-v}}\,\mathrm{d}v\\
&=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\frac{\Gamma(k+1)\Gamma(1/2)}{\Gamma(k+3/2)}\\
&=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\frac{k!\sqrt\pi}{(k+1/2)k!\binom{2k}{k}\frac{\sqrt\pi}{4^k}}\\
&=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\frac{2^{2k+1}}{(2k+1)\binom{2k}{k}}
\end{align}
$$
A: 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. 
