Integrate $\int_0^\pi e^{-ik\cos\theta}\sin^2{\theta} \, \mathrm{d}\theta$ Does anyone have some good method to integrate
$$\int_0^\pi e^{-ik\cos\theta}\sin^2{\theta} \, \mathrm{d}\theta$$

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:
$$\begin{align}
\mathcal{I}
&=\int_{0}^{\pi}e^{-ik\cos{\theta}}\sin^2{\theta}\,\mathrm{d}\theta\\
&=\int_{0}^{\pi}e^{-ik\cos{\theta}}\sqrt{1-\cos^2{\theta}}\,\sin{\theta}\,\mathrm{d}\theta\\
&=\int_{-1}^{1}e^{ikx}\sqrt{1-x^2}\,\mathrm{d}x\\
&=\int_{0}^{1}e^{ikx}\sqrt{1-x^2}\,\mathrm{d}x+\int_{-1}^{0}e^{ikx}\sqrt{1-x^2}\,\mathrm{d}x\\
&=\int_{0}^{1}e^{ikx}\sqrt{1-x^2}\,\mathrm{d}x+\int_{0}^{1}e^{-ikx}\sqrt{1-x^2}\,\mathrm{d}x\\
&=2\int_{0}^{1}\frac{e^{ikx}+e^{-ikx}}{2}\sqrt{1-x^2}\,\mathrm{d}x\\
&=2\int_{0}^{1}\cos{\left(kx\right)}\sqrt{1-x^2}\,\mathrm{d}x\\
&=\frac{\pi\,J_{1}{\left(k\right)}}{k}.\\
\end{align}$$
