Find $\int \limits_0^\pi \sin(\sin(x))\sin(x)\mathrm dx$ Compute $\displaystyle \int \limits_0^\pi \sin(\sin(x))\sin(x)\mathrm dx$.
I have no idea how to integrate of this. I do need some help. Thanks
 A: 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 $$\int_0^\pi \sin(\sin x) \sin x \, dx = \pi \sum_{n=0}^{\infty}\frac{(-1)^n}{2^{2n+1} n! (n+1)!} = \pi J_1(1) \approx 1.38246$$
A: Hint: What happens when we try the substitution $u = \sin{x}$?
A: This reminds me of 
$ \frac{1}{\pi}\int_{0}^{\pi} \sin(\cos(\sin(\cos( x)))) dx   $ where we iterate the expression in the integrand, replacing $x$ by $ \sin(\cos( x))  $ and  which goes to   $0.6948196907307875\ldots$ (a fixed point of  $\sin(\cos(x_0)) = x_0$).
