Question:
Find out $\displaystyle{\int e^{\cos x}~dx}$.
My Attempt:
Let $\cos x = y$. Hence $-\sin x\ dx = dy$ or $$dx = \displaystyle{\frac{-dy}{\sin x}=\frac{-dy}{\sqrt{1-\cos^2x}}=\frac{-dy}{\sqrt{1-y^2}}}$$ So
$$\begin{align}\int e^{\cos x}~dx &= \int e^y\left(\frac{-dy}{\sqrt{1-y^2}}\right)\\
&=-\int\frac{e^y}{\sqrt{1-y^2}}~dy
\end{align}$$
This integral is one I can't solve. I have been trying to do it for the last two days, but can't get success. I can't do it by parts because the new integral thus formed will be even more difficult to solve. I can't find out any substitution that I can make in this integral to make it simpler. Please help me solve it. Is the problem with my first substitution $y=\cos x$ or is there any other way to solve the integral $\displaystyle{\int\frac{e^y}{\sqrt{1-y^2}}~dy}$?
Best Answer
See also https://math.stackexchange.com/a/117545/442
Although this indefinite integral has no known closed form, certain definite integrals are known... $$ \int_0^\pi e^{\cos x}\;dx = \pi\;I_0(1) , $$ where $I_0$ is a modified Bessel function