Find $\int_{0}^{\pi} \sin(A\sin(x))\sin(x) \,dx$ ($A$ is constant) How does one evaluate the integral
$$\int_{0}^{\pi} \sin(A\sin(x))\sin(x) \,dx $$ where $A$ is a constant?
Thanks to Perry Iverson，Steven Stadnicki，Jack D'Aurizio for my former question by Taylor series. However,  multiplying by a constant A, can it still be solved in the same way? Thank you.
 A: $$\int^\pi_0\sin(A\sin x)\sin xdx$$
Now,
$$\sin(A\sin x)=\sum^\infty_{k=0}\dfrac{(-1)^kA^{2k+1}}{(2k+1)!}\sin^{2k+1}x$$
Thus,
$$\sin(A\sin x)\sin x=\sum^\infty_{k=0}\dfrac{(-1)^kA^{2k+1}}{(2k+1)!}\sin^{2k+2}x$$
The integral thus is
$$\int^\pi_0\sin(A\sin x)\sin xdx=\sum^\infty_{k=0}\dfrac{(-1)^kA^{2k+1}}{(2k+1)!}\int\sin^{2k+2}xdx$$
$$\forall n\in\mathbb{Z}_{\geq0}, \int^\pi_0\sin(A\sin x)\sin xdx=\pi\sum^\infty_{k=0}\dfrac{(-1)^kA^{2k+1}}{2^{2k+1}k!(k+1)!}$$
From the accepted answer here. You can use Bessel functions here too to get $$\forall n\in\mathbb{Z}_{\geq0}, \int^\pi_0\sin(A\sin x)\sin xdx=\pi\sum^\infty_{k=0}\dfrac{(-1)^kA^{2k+1}}{2^{2k+1}k!(k+1)!}=\pi J_1(A)$$
Where $J_n$ is the Bessel function of the first kind.
A: Here is an interesting approach...We first show a definition for the Bessel Function of the first kind:
$$ J_n(x)=\frac{1}{\pi}\int_0^\pi\cos[nu-x\sin(u)]du  $$
Now, setting $n=0$, and realizing that $\cos(-u)=\cos(u)$, we get:
$$ \pi J_0(A) = \int_0^\pi \cos[A\sin(u)]du$$
Next, we take the partial derivative of both sides with respect to A:
$$\frac{\partial}{\partial A} \pi J_0(A) = \int_0^\pi \frac{\partial}{\partial A} \cos[A\sin(u)]du$$
Remember that $J_0'(x)=-J_1(x)$.
$$ -\pi J_1(A)= -\int_0^\pi \sin[A\sin(u)]\sin(u)du $$
Now, we are left with the original integral!
$$ \color{blue}{\int_0^\pi \sin[A\sin(x)]\sin(x)dx=\pi J_1(A)}  $$
