I encountered the above when dealing with the Bessel functions of the first kind, $J_n(z)$, specifically $n=0$. Using the differential-equation definition of the Bessel function, I obtained the above question's equivalence to $$I(z)=\int_0^\pi e^{i z \cos(\theta)}( \cos(\theta)+iz\sin^2(\theta))d \theta =0.$$
How is this proved?
I think the series definition is best. The series definition of the Bessel function is
$$J_{\nu}(z) = \sum_{n=0}^{\infty}\frac{(-1)^n}{\Gamma(n+\nu+1)\,n!}\left(\frac{x}{2}\right)^{2n+\nu}.$$
We will expand $e^{iz\cos(\theta)}$ as a power series as follows:
$$e^{iz\cos(\theta)} = \sum_{n=0}^{\infty}\frac{1}{n!}(iz\cos(\theta))^n.$$
So integrating term by term (which we can do since our series converges absolutely and uniformly everywhere) we have
$$\frac{1}{\pi}\int_0^{\pi}e^{iz\cos(\theta)}d\theta = \frac{1}{\pi}\int_0^{\pi}\sum_{n=0}^{\infty}\frac{1}{n!}(iz\cos(\theta))^nd\theta = \frac{1}{\pi}\sum_{n=0}^{\infty}\frac{1}{n!}i^nz^n\int_0^{\pi}\cos^n(\theta)d\theta.$$
Now we should make use of the identity (which you can prove by induction)
$$\int_0^{\pi}\cos^n(\theta)d\theta = \begin{cases}0 & n \text{ odd}\\\frac{(2n)!\pi}{2^{2n}(n!)^2} & n \text{ even}\end{cases}. $$
Then we have
$$\frac{1}{\pi}\int_0^{\pi}e^{iz\cos(\theta)}d\theta = \frac{1}{\pi}\sum_{n=0}^{\infty}\frac{1}{(2n)!}i^{2n}z^{2n}\frac{(2n)!\pi}{2^{2n}(n!)^2} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(n!)^2}\left(\frac{z}{2}\right)^{2n}. $$
This is exactly what we wanted to show (noting that $\Gamma(n+1) = n!$).