An issue with the substitution $u=\sin x$ I'm embarrassed to ask this question, but what's the error in the following evaluation?
$$\int_{0}^{\pi} \sin (\sin x) \ dx = \int_{0}^{0} \frac{\sin u}{\sqrt{1-u^{2}}} \ du = 0$$  
Is it that $x = \arcsin u$ only for $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$?
 A: Your $u$-substitutions should be injective on their interval of evaluation. Otherwise, you risk running into this sort of issue.
Now, if you want to use $u=\sin x$, then $$\frac{du}{dx}=\cos x=\begin{cases}|\cos x|=\sqrt{1-u^2} & 0\le x\le \frac\pi2\\-|\cos x|=-\sqrt{1-u^2} & \frac\pi2\le x\le\pi,\end{cases}$$ so $$\begin{align}\int_0^\pi\sin(\sin x)\,dx &= \int_0^{\pi/2}\sin(\sin x)\,dx+\int_{\pi/2}^\pi\sin(\sin x)\\ &= \int_0^{\pi/2}\sin(\sin x)\,dx-\int_\pi^{\pi/2}\sin(\sin x)\,dx\\ &= \int_0^1\frac{\sin u}{\sqrt{1-u^2}}\,du-\int_0^1\frac{\sin u}{-\sqrt{1-u^2}}\,du\\ &= 2\int_0^1\frac{\sin u}{\sqrt{1-u^2}}\,du.\end{align}$$ Alternately, you could note that $\sin(\pi-x)=\sin x$, so $$\begin{align}\int_0^\pi\sin(\sin x)\,dx &= \int_0^{\pi/2}\sin(\sin x)\,dx+\int_{\pi/2}^\pi\sin(\sin x)\\ &= \int_0^{\pi/2}\sin(\sin x)\,dx-\int_\pi^{\pi/2}\sin(\sin x)\,dx\\ &= \int_0^{\pi/2}\sin(\sin x)\,dx-\int_\pi^{\pi/2}\sin(\sin(\pi-x))\,dx\\ &= \int_0^{\pi/2}\sin(\sin x)\,dx-\int_0^{\pi/2}\sin(\sin x)\frac{d(\pi-x)}{dx}\,dx\\ &= 2\int_0^{\pi/2}\sin(\sin x)\,dx,\end{align}$$ at which point you can use your $u$-substitution without fear, since the sine function is injective on $[0,\pi/2]$.
I'm afraid that integral isn't going to have a nice elementary evaluation. W|A gives a solution in terms of the Struve function.
A: Your problem is not that $\sin x$ is not one-to-one on $0\le x\le \pi$, but that you took $\cos x$ as $\sqrt{1-\sin^2x}$ on $0\le x\le \pi$, and this is wrong because on $\pi/2\le x\le \pi$, $$\cos x=-\sqrt{1-\sin^2x}.$$
