I'm self studying contour integration and I found a problem I can't solve. Prove for a real number $k$ that
$$\int_{0}^{\pi} e^{k \cos(\theta)} \cos(k \sin(\theta))d \theta = \pi$$
I tried doing the change of variable $x = \cos \theta$ which gets me the integal
$$ \int_{-1}^{1} e^{kx} \frac{cos(k\sqrt{1-x^2})}{\sqrt{1-x^2}}dx $$ but to no avail. I was thinking to use a half circle contour but I haven't found an appropiate function. Any help will be appreciated.
A good start may be to consider the integral $\int_\gamma \frac{e^z}{z} dz$, where $\gamma$ denotes a circle of radius $k$ centered at the origin. By Cauchy's residue theorem, this integral has value $2 \pi i$. Of course, under the change of variable $z = ke^{it}$, the above becomes $$i \int_0^{2 \pi} e^{k e^{it}} dt$$ So that $$\int_{0}^{2 \pi} e^{ke^{i t}} dt = 2 \pi$$ Now, recall that $k e^{it} = k\cos (t) + i k\sin (t)$, so that $$e^{ke^{it}} = e^{k\cos (t) } \big( \cos (k\sin (t) ) + i \sin (k\sin (t)) \big)$$ Integrating and taking real parts, we see: $$\int_0^{2 \pi} e^{k \cos (t)} \cos (k \sin (t)) dt = 2 \pi$$ And, to get the integral from $0$ to $\pi$, simply divide this integral by $2$ since $\sin$ and $\cos$ attain the same values on $[\pi , 2 \pi]$ as on $[0 , \pi]$. Thus, $$\int_0^\pi e^{k \cos (t)} \cos (k \sin (t)) dt = \pi$$ As desired.