Since the only singularity is at $z=0$, we get that
$$
\frac{e^{\alpha z}}{z}=\frac{1+\alpha z+\frac12\alpha^2z^2+\frac16\alpha^3z^3+\dots}{z}\tag{1}
$$
Thus, as long as $C$ circles the origin once clockwise,
$$
\int_{C}\frac{e^{\alpha z}}{z}\mathrm{d}z=2\pi i\tag{2}
$$
Notice that with $z=e^{it}=\cos(t)+i\sin(t)$,
$$
\begin{align}
\int_C \frac{e^{\alpha z}}{z}\,\mathrm{d}z
&=\int_{-\pi}^\pi e^{\alpha(\cos(t)+i\sin(t))}\,i\,\mathrm{d}t\\
&=i\int_{-\pi}^\pi e^{\alpha\cos(t)}(\cos(\alpha\sin(t))+i\sin(\alpha\sin(t)))\,\mathrm{d}t\\
&=i\int_{-\pi}^\pi e^{\alpha\cos(t)}\cos(\alpha\sin(t))\,\mathrm{d}t\\
&=2i\int_0^\pi e^{\alpha\cos(t)}\cos(\alpha\sin(t))\,\mathrm{d}t\tag{3}
\end{align}
$$
Combining $(2)$ and $(3)$ yields
$$
\int_0^\pi e^{\alpha\cos(t)}\cos(\alpha\sin(t))\,\mathrm{d}t=\pi
$$