Evaluating this complicated integral using complex analysis I am trying to evaluate this integral:
$$\boxed{\int_{-\pi}^\pi\sin(2\cos\theta)\cos((2m+1)\theta)\,d\theta}$$
where $m\in\mathbb{N}$, using complex analysis methods.
What I have done is to find the residue of $\sin(z+\frac 1z)$:
Since $$\sin(z+\frac 1z)=\sum_{k=0}^\infty\frac{(-1)^k(z+z^{-1})^{2k+1}}{(2k+1)!}$$, and using the Binomial theorem, we have $$(z+z^{-1})^{2k+1}=\sum_{r=0}^{2k+1}{{2k+1}\choose r}z^{2k+1-2r}$$.
Combining these two facts we have that the residue (coefficient of $z^{-1}$) is:
$$\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}{{2k+1}\choose{k+1}}$$.
After this I am kind of stuck.
I am aware of the standard technique of substituting $\cos\theta=\frac{z+z^{-1}}{2}$, $d\theta=\frac{dz}{iz}$, etc, but I am not sure what to do with the $\cos((2m+1)\theta)$.
Thanks for any help.
By the way, if it helps, the answer I am supposed to get is $$\boxed{2\pi\sum_{k=m}^\infty\frac{(-1)^k}{(2k+1)!}{{2k+1}\choose {k-m}}}$$.
Update: Using David Holden's excellent hint, I proceded to find the residue of the integrand. 
So the integral becomes $\frac{1}{2i}\int_C\sin(z+z^{-1})(z^{2m}+z^{-2m-2})\,dz$.
I find the residue of the integrand to be 
$$\sum_{k=m}^\infty \frac{(-1)^k}{(2k+1)!}{{2k+1}\choose{k+m+1}}+\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}{{2k+1}\choose{k-m}}$$
I may have made a mistake along the way, I can't get to the final answer.
 A: Consider that $\cos{\left [ (2 m+1) \theta \right ]} = \operatorname{Re}{\left [e^{i (2 m+1) \theta}\right ]} $.  Then we consider the integral
$$-i \oint_{|z|=1} dz \, \sin{\left ( z+\frac1{z} \right )} z^{2 m}$$
which is equal to $i 2 \pi$ times the sum of the residues of the integrand at $z=0$.  To evaluate the residues and their sum, we Taylor expand the sine term to get, as the integrand,
$$-i \sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k+1)!} \left ( z+\frac1{z} \right )^{2 k+1} z^{2 m} $$
The coefficient of $z^{-1}$ in the $k$th term of the expansion is the coefficient of the $k+m+1$th term.  (You can work this out for successive values of $m$.)  Thus, the integral is equal to
$$2 \pi \sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k+1)!} \binom{2 k+1}{k+m+1} = 2 \pi \sum_{k=0}^{\infty} \frac{(-1)^k}{(k-m)! (k+m+1)!}$$
To evaluate the sum, note that the terms for $k \lt m$ are zero.  Then we can shift the index of the sum to get
$$ 2 \pi \sum_{k=m}^{\infty} \frac{(-1)^k}{(k-m)! (k+m+1)!} =  2 \pi (-1)^m \sum_{k=0}^{\infty} \frac{(-1)^k}{k! (k+2 m+1)!}$$
The latter sum is recognizable as a Bessel function evaluated at $z=2$.  Thus,

$$\int_{-\pi}^{\pi} d\theta \, \sin{(2 \cos{\theta})} \cos{\left [ (2 m+1) \theta \right ]} = 2 \pi (-1)^m J_{2 m+1}(2)$$

