Originally Posted by
Random Variable It's not as nice as Opalg's solution, but how about the following:
Let $\displaystyle I(a) = \int_{0}^{\pi} e^{a \cos x} \cos(a \sin x) \ dx = Re \Big(\int_{0}^{\pi} e^{ae^{ix}} \ dx \Big) $
now assuming it's OK to differentiate inside of the integral with respect to $\displaystyle a$
$\displaystyle I'(a) = Re \Big(\int_{0}^{\pi} e^{ix} e^{ae^{ix}} \ dx \Big) = Re \Big( \frac{1}{ai} \int_{a}^{-a}e^{u} \ du \Big) = Re \Big( \frac{i}{a} (e^{a}-e^{-a}) \Big) =0 $
so $\displaystyle I(a) = C$
which means that $\displaystyle I$ is independent of the value of $\displaystyle a$
to find $\displaystyle C $ let $\displaystyle a=0$
then $\displaystyle I(0)= C = \int^{\pi}_{0} dx = \pi $
and $\displaystyle \int_{0}^{\pi} e^{\cos x} \cos(\sin x) \ dx = I(1) = \pi $