Showing $\int_{0}^{2 \pi} \cos(\cos \theta)\cosh(\sin \theta) d \theta = 2\pi$

Question

I attempted using the substitutions $$ \cos( \theta )=\frac{e^{i\theta}+e^{-i \theta}}{2}$$ and $$\cosh(\theta)=\frac{e^{\theta}+e^{-\theta}}{2}$$ but ended up with a messy equation involving a lot of exponential terms to the power of exponential terms, is there an easier way to go? Help would be appreciated thanks!

Answer 1

\begin{align} \int_0^{2\pi}\cos(\cos\theta)\cosh(\sin\theta)\,d\theta &= \operatorname{Re}\int_0^{2\pi}\cos\left(e^{i\theta}\right)\,d\theta\\ &=\operatorname{Re}\frac{1}{i}\oint_{|z|=1}\frac{\cos z}{z}\,d\theta\\ &=\operatorname{Re}2\pi\text{ Res }\left(\frac{\cos z}{z}, 0\right)\\ &=2\pi\operatorname{Re}\lim\limits_{z\rightarrow 0}\cos z\\ &=2\pi \end{align}

Answer 2

Hint. Notice that, for $\theta \in\mathbb{R}$, $$ \cos(\cos \theta)\cosh(\sin \theta)=\text{Re}\left[\cos\left(e^{i\theta} \right)\right] $$ and apply Cauchy's integral theorem to $$ \int_{|z|=1}\frac{\cos(z)}{i\,z}\,dz. $$