Integration of exponential functions and cosine function I am trying to solve the following equation;
$$\int_{-1}^{1}e^{i(x+a\cos x)} \, \mathrm{d}(\cos x)$$ or $$\int_{0}^{\pi}e^{i(x+a\cos x)} \sin x \, \mathrm{d}x$$
I tried this in Wolfram Alpha, but it says that answer cannot be obtained.
 A: Let $F(a)$ be defined as
$$F(a)=\int_0^{\pi}e^{i(x+a\cos x)}\sin x\,dx \tag 1$$
Now, it is easy to show that the real part of $F$ is identically $0$.  To do this, we write the real part of $(1)$ as 
$$\text{Re}\lbrace F(a)\rbrace=\int_0^{\pi/2}\cos(x+a\cos x)\sin x\,dx+\int_{\pi/2}^{\pi}\cos(x+a\cos x)\sin x\,dx \tag 2$$
Then, enforcing the substitution $x\to \pi -x$ in the second integral on the right-hand side of $(2)$ reveals
$$\begin{align}
\text{Re}\lbrace F(a)\rbrace&=\int_0^{\pi/2}\cos(x+a\cos x)\sin x\,dx+\int_{0}^{\pi/2}\cos(\pi-x+a\cos (\pi-x))\sin (\pi-x)\,dx\\\\
& =\int_0^{\pi/2}\cos(x+a\cos x)\sin x\,dx-\int_0^{\pi/2}\cos(x+a\cos x)\sin x\,dx\\\\
&=0 \tag 3
\end{align}$$
as was to be shown.  We can write, therefore, $F(a)$ as 
$$\begin{align}
F(a)&=i\int_0^{\pi}\sin(x+a\cos x)\sin x\,dx \\\\
&=i\int_0^{\pi}\left(\sin^2(x)\cos(a\cos x)+\sin(x)\cos(x)\sin(a\cos x)\,\right)dx\tag 4
\end{align}$$
where we arrived at the right-hand side of $(4)$ using a parallel development that led to $(3)$.

The second integral on the right-hand side of $(4)$ is easily evaluated.  Substituting $x\to \cos(x)$ and integrating by parts reveals that 
$$\begin{align}
i\int_0^{\pi}\sin(x)\cos(x)\sin(a\cos x)\,dx&=i2\int_{0}^1 x\sin(ax)\,dx\\\\
&=i2\frac{-a\cos(a)+\sin(a)}{a^2} \tag 5
\end{align}$$
Using $(5)$ in $(4)$ reduces $F(a)$ to 
$$\begin{align}
F(a)&=i\int_0^{\pi}\left(\sin^2(x)\cos(a\cos x)\right)\,dx\\\\
&+i2\frac{-a\cos(a)+\sin(a)}{a^2} \tag 6
\end{align}$$

In order to evaluate the integral in $(6)$ we will enforce two successive substitutions.  First, we let $a\cos x\to x$.  This yields
$$i\int_0^{\pi}\left(\sin^2(x)\cos(a\cos x)\right)=\frac{i2}{a}\int_0^a
\sqrt{1-\left(\frac{u}{a}\right)^2} \cos (x)\,dx$$
where we exploited the fact that the integrand is an even function.  Next, we enforce the substitution $x\to a\sin x$ which reveals 
$$\begin{align}
i\int_0^{\pi}\left(\sin^2(x)\cos(a\cos x)\right)&=i\int_0^{\pi}\cos^2(x)\cos(a\sin x)\,dx \tag 7\\\\
&=i\int_0^{\pi}\cos(a\sin x)\,dx-i\int_0^{\pi}\sin^2(x)\cos(a\sin x)\,dx\\\\
&=i\pi J_0(a)+i\pi J_1'(a)\\\\
&=i\pi\frac{J_1(a)}{a} \tag 8
\end{align}$$
where $J_0(a$) and $J_1(a)$ are the Bessel Function of the first kind and order zero and one, respectively, while $J_1'(a)$ is the first derivative of $J_1$.  In deriving $(8)$ we relied on Equations $(B5)$, $(B6)$, and $(B8)$ in the section on Bessel Functions.
Aside, we exploited the symmetry of $\cos^2(x)\cos(a\sin x)$ around $\pi/2$ to arrive at $(7)$.

FINAL RESULT
Putting $(6)$ and $(8)$ together finally yields
$$\bbox[5px,border:2px solid #C0A000]{F(a)=i2\frac{-a\cos(a)+\sin(a)}{a^2}+i\pi\frac{J_1(a)}{a}} $$
which agrees with the empirically obtained hypothesis of @ClaudeLeibovici and @JoshBroadhurst!

BESSEL FUNCTION ASIDE
We note an integral representation for the first kind Bessel Function of order $n$
$$\begin{align}
J_n(a)&=\frac{1}{\pi}\int_0^{\pi}\cos(nx-a\sin x)dx \\\\
&=\frac{1}{\pi}\int_0^{\pi} \left(\cos (nx)\cos(a\sin x)+\sin(nx)\sin(a\sin x)\right) \,dx \tag{B1}
\end{align}$$
along with recurrence relationships
$$J_n(a)=\frac{a}{2n}\left(J_{n-1}(a)+J_{n+1}(a)\right) \tag{B2}$$
and
$$J_n'(a)=\frac12
\begin{cases}
J_{n-1}(a)-J_{n+1}(a),&n\ne 0\\\\
-J_1(a),&n = 0 \tag{B4}
\end{cases}$$
Setting $n=0$ in $(B1)$ yields
$$\bbox[5px,border:2px solid #C0A000]{J_0(a)=\frac{1}{\pi}\int_0^{\pi} \left(\cos(a\sin x)\right) \,dx} \tag{B5}$$
Taking the two derivatives of $J_0(a)$ and multiplying by $-1$ and using $(B4)$ yields
$$\bbox[5px,border:2px solid #C0A000]{J_1'(a)=-\frac{1}{\pi}\int_0^{\pi} \sin^2(x)\cos(a\sin x)\,dx}\tag{B6}$$
Note that using $(B3)$ and $(B4)$, we can eliminate $J_{n+1}$ and write $J_n$ as
$$J_n(a)=\frac{a}{n}\left(J_{n-1}(a)-J_n'(a)\right) \tag{B7}$$
Setting $n=1$ in $(B7)$ and rearranging we obtain
$$\bbox[5px,border:2px solid #C0A000]{\frac{J_1(a)}{a}=J_0(a)+J_1'(a)} \tag{B8}$$

A: If you leave out the (what I assume is a) constant $a$ in your original problem and type it into WolframAlpha, it tells you that the solution involves the Bessel function of the first kind ($J_1$) and gives the following answer
$$ (\pi J_1(1)-2 \cos(1)+2 \sin(1))i \approx 1.9848i$$ 
If you set $a=2$ as the input to WolframAlpha this gives
$$ \frac{1}{2} (\pi J_1(2)-2 \cos(2)+\sin(2))i \approx 1.7767i$$
Again with $a=3$ as the input gives
$$ \frac{1}{9} (3 \pi J_1(3)-6 \cos(3)+2 \sin(3))i \approx 1.0464i$$
Perhaps you can use a few more values for $a$ and generalize a pattern. 
The input I'm using looks like:
integrate from 0 to pi: e^(i (x + 3 cosx)) sinx dx

Hopefully this is sufficient for your needs.
A: In the same spirit as Josh Broadhurst's answer, you should arrive to $$\int_{0}^{\pi}e^{i(x+a\cos(x))} \sin(x)\,dx=\frac{\pi  a J_1(a)-2 a \cos (a)+ 2 \sin (a)}{a^2}\,i$$  I do not find  any way to compute the antiderivative itself.
