integration - Calculating the following integral using complex analysis: $int_{0}^{pi}e^{acos(theta)}cos(asin(theta)), dtheta$

I am trying to solve a question from my complex analysis test that
I didn't manage to do during the test in order to practice for the
next exam.

The problem is as follows:

Calculate $\int_{0}^{\pi}e^{a\cos(\theta)}\cos(a\sin(\theta))\, d\theta$

Where the method used should be using complex analysis.

What I tried:

I have noticed $$e^{a\cos(\theta)}\cos(a\sin(\theta))=Re(e^{acis(\theta)})$$
where $cis(\theta):=\cos(\theta)+i\sin(\theta)$

Hence the integral is $$\int_{0}^{\pi}Re(e^{a(\cos(\theta)+i\sin(\theta))}\, d\theta$$

I did the change of variables: $z=ae^{i\theta}$, $dz=iae^{i\theta}\, d\theta\implies d\theta=\frac{dz}{iz}$.

When $\theta=0$ we have that $z=a$ and when $\theta=\pi$ we get
$z=-ia$

and so I wrote that the integral is

$$\int_{a}^{-ia}Re(e^{z})\,\frac{dz}{iz}$$

Now I don't understand how I got $e^{z}$, which seems wrong to me,
but what the checker actually marked in red and wrote a question mark
by were the integration limits: $a,-ia$ .

Can someone please help me understand what was wrong with what the
integration limits, and how to actually solve this integral ?

Any help is greatly appreciated!

Answer

As noted in Emmet´s comment, there is an error in your computation: when $\theta=\pi$ $z=-a$, not $-i\,a$ ($e^{\pi i}=-1$).

The computation becomes easier by noting that the integrand is even, so that it is one half of the integral between $0$ and $2\,\pi$. After the change of variables the integral becomes a line integral along the unit circle, and you can apply Cauchy's theorem or the calculus of residues.

I will work backwards. Consider the integral
$$\int_{|z|=1}\frac{e^{az}}{i\,z}\,dz.$$
By Cauchy's theorem its value is $2\,\pi$. Now let $z=e^{it}$. Then
$$\begin{align*} \int_{|z|=1}\frac{e^{az}}{i\,z}\,dz&=\int_0^{2\pi}e^{ae^{it}}\frac{i\,e^{it}\,dt}{i\,e^{it}}\\ &=\int_0^{2\pi}e^{a(\cos t+i\sin t)}dt\\ &=\int_0^{2\pi}e^{a\cos t}\cos(a\sin t)\,dt+i\int_0^{2\pi}e^{a\cos t}\sin(a\sin t)\,dt. \end{align*}$$
Finally, we get
$$\int_0^{\pi}e^{a\cos t}\cos(a\sin t)\,dt=\pi.$$