Definite trig integral [duplicate]

Question

How do I evaluate:

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

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

Answer 1

This is not a closed form (I don't know if one exists in elementary functions), but this series converges pretty fast: $$ \begin{align} \int_0^\pi\sin(\sin(x))\,\mathrm{d}x &=2\int_0^1\frac{\sin(u)}{\sqrt{1-u^2}}\,\mathrm{d}u\\ &=2\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\int_0^1\frac{u^{2k+1}}{\sqrt{1-u^2}}\,\mathrm{d}u\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\int_0^1\frac{v^k}{\sqrt{1-v}}\,\mathrm{d}v\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\frac{\Gamma(k+1)\Gamma(1/2)}{\Gamma(k+3/2)}\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\frac{k!\sqrt\pi}{(k+1/2)k!\binom{2k}{k}\frac{\sqrt\pi}{4^k}}\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\frac{2^{2k+1}}{(2k+1)\binom{2k}{k}} \end{align} $$

Answer 2

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.