Find $\int \limits_0^\pi \sin(\sin(x))\sin(x)\mathrm dx$

Here's a solution using Taylor Series:

$$\sin(\sin x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}(\sin x)^{2n+1}$$

$$\sin(\sin x) \sin x = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}(\sin x)^{2n+2}$$

$$\int_0^\pi \sin(\sin x) \sin x \, dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\int_0^\pi(\sin x)^{2n+2} dx$$

Note that $\int_0^\pi(\sin x)^{2n+2} dx = \pi \frac{(2n+1)(2n-1)(2n-3)\cdots 3 \cdot 1}{(2n+2)(2n)(2n-2)\cdots 4 \cdot 2} = \pi \frac{(2n+1)!!}{(2n+2)!!}$ for any non-negative integer $n$. You can use the following reduction formula to prove it:

$$\int \sin^n x \, dx = - \frac{1}{n} \sin^{n-1} x \cos^{n-1} x + \frac{n-1}{n} \int \sin^{n-2} x \, dx$$

So we have:

$$\begin{align} \int_0^\pi \sin(\sin x) \sin x \, dx &= \pi \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)} \frac{(2n+1)!!}{(2n+2)!!} \\ &= \pi \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!!(2n+2)!!} \\ &= \pi \sum_{n=0}^{\infty}\frac{(-1)^n}{(2^n n!) (2^{n+1} (n+1)!)} \\ &= \pi \sum_{n=0}^{\infty}\frac{(-1)^n}{2^{2n+1} n! (n+1)!} \\ \end{align}$$

Now you can note that the summation here is exactly the definition of $J_1(1)$, where $J_\alpha(x)$ is the Bessel function of the first kind:

http://en.wikipedia.org/wiki/Bessel_function

So $$\int_0^\pi \sin(\sin x) \sin x \, dx = \pi \sum_{n=0}^{\infty}\frac{(-1)^n}{2^{2n+1} n! (n+1)!} = \pi J_1(1) \approx 1.38246$$