Transform complex exponential integral to real Question: Transform
$J_n (x)=\frac{1}{2 \pi} \int_{\theta=-\pi}^\pi e^{i(n\theta-x\sin\theta)}$
On answer sheet:
$J_n (x)=\frac{1}{\pi} \int_{\theta=0}^\pi \cos(n\theta-x\sin\theta)$
But how? I am not very good at complex analysis, thanks for any help!

By definition, the real part:
$$e^{i(n\theta-x\sin\theta)}=\cos(n\theta-x\sin\theta)+i\sin(n\theta-x\sin\theta)\implies$$
$$\implies \text{Re}\,e^{i(n\theta-x\sin\theta)}=\cos(n\theta-x\sin\theta)$$
and as this last one is an even function, we get
$$\frac1{2\pi}\int_{-\pi}^\pi\cos(n\theta-x\sin\theta)=2\left(\frac1{2\pi}\int_0^\pi\cos(n\theta-x\sin\theta)\right)=\frac1{\pi}\int_0^\pi\cos(n\theta-x\sin\theta)$$
