Integrate $\int_0^\pi e^{-ik\cos\theta}\sin^2{\theta} \, \mathrm{d}\theta$ Does anyone have some good method to integrate
$$\int_0^\pi e^{-ik\cos\theta}\sin^2{\theta} \, \mathrm{d}\theta$$
 A: As a first step, you might substitute $x=-\cos{\theta}$. Next, a little algebra puts the integral in a form recognizable as a common integral representation for the Bessel function of the first kind of order one:
$$\begin{align}
\mathcal{I}
&=\int_{0}^{\pi}e^{-ik\cos{\theta}}\sin^2{\theta}\,\mathrm{d}\theta\\
&=\int_{0}^{\pi}e^{-ik\cos{\theta}}\sqrt{1-\cos^2{\theta}}\,\sin{\theta}\,\mathrm{d}\theta\\
&=\int_{-1}^{1}e^{ikx}\sqrt{1-x^2}\,\mathrm{d}x\\
&=\int_{0}^{1}e^{ikx}\sqrt{1-x^2}\,\mathrm{d}x+\int_{-1}^{0}e^{ikx}\sqrt{1-x^2}\,\mathrm{d}x\\
&=\int_{0}^{1}e^{ikx}\sqrt{1-x^2}\,\mathrm{d}x+\int_{0}^{1}e^{-ikx}\sqrt{1-x^2}\,\mathrm{d}x\\
&=2\int_{0}^{1}\frac{e^{ikx}+e^{-ikx}}{2}\sqrt{1-x^2}\,\mathrm{d}x\\
&=2\int_{0}^{1}\cos{\left(kx\right)}\sqrt{1-x^2}\,\mathrm{d}x\\
&=\frac{\pi\,J_{1}{\left(k\right)}}{k}.\\
\end{align}$$
A: David H's answer is perfectly fine. 
Anyway, if you are not confident with Bessel functions, you can just notice that:
$$ I = \int_{0}^{\pi}e^{-ik\cos\theta}\sin^2\theta\,d\theta = 2\int_{0}^{\pi/2}\cos(k\cos\theta)\sin^2\theta\,d\theta\\=\frac{2}{k}\int_{0}^{\pi/2}\sin(k\cos\theta)\cos\theta\,d\theta $$
where the last equality follows from integration by parts. Since:
$$\int_{0}^{\pi/2}\cos^{2m}\theta\,d\theta=\frac{\pi}{4^{m}}\binom{2m}{m},$$
exploiting the Taylor series of the sine function we get:
$$ I = \frac{2}{k}\sum_{n=0}^{+\infty}(-1)^{n}\frac{k^{2n+1}}{(2n+1)!}\int_{0}^{\pi/2}\cos^{2n+2}\theta\,d\theta=\frac{\pi}{4}\sum_{n=0}^{+\infty}\frac{(-1)^n k^{2n}}{4^n(2n+1)!}\binom{2n+2}{n+1}$$
or just:
$$\color{purple}{ I = \frac{\pi}{2}\sum_{n=0}^{+\infty}\frac{(-1)^n}{n!(n+1)!}\left(\frac{k^2}{4}\right)^n} $$
that is a pretty fast converging series.
