Integrating Exp[Cos^2] I'm trying to integrate something of this form:
$\int_0^se^{\cos^2x}~dx$
I looked in different tables and tried with Mathematica but without success... Could you give me the solution (if obvious) or advices on how to proceed?
 A: $\int_0^se^{\cos^2x}~dx=\int_0^s\sum\limits_{n=0}^\infty\dfrac{\cos^{2n}x}{n!}dx=\int_0^s\left(1+\sum\limits_{n=1}^\infty\dfrac{\cos^{2n}x}{n!}\right)dx$
For $\int\cos^{2n}x~dx$ , where $n$ is any natural number,
$\int\cos^{2n}x~dx=\dfrac{(2n)!x}{4^n(n!)^2}+\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin x~\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$
This result can be done by successive integration by parts, e.g. as shown as http://hk.knowledge.yahoo.com/question/question?qid=7012022000808
$\therefore\int_0^s\left(1+\sum\limits_{n=1}^\infty\dfrac{\cos^{2n}x}{n!}\right)dx$
$=\left[x+\sum\limits_{n=1}^\infty\dfrac{(2n)!x}{4^n(n!)^3}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin x~\cos^{2k-1}x}{4^{n-k+1}(n!)^3(2k-1)!}\right]_0^s$
$=\left[\sum\limits_{n=0}^\infty\dfrac{(2n)!x}{4^n(n!)^3}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin x~\cos^{2k-1}x}{4^{n-k+1}(n!)^3(2k-1)!}\right]_0^s$
$=\sum\limits_{n=0}^\infty\dfrac{(2n)!s}{4^n(n!)^3}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin s~\cos^{2k-1}s}{4^{n-k+1}(n!)^3(2k-1)!}$
