Integration of hyperbolic function $$\int \tanh \left ( \cos x \right )\mathrm{d}x$$
Is this analytically Solvable?
 A: Since the radius of convergence of maclaurin series of $\tanh x$ is $\dfrac{\pi}{2}$ , the following procedure holds for at least all real numbers of $x$ :
$\int\tanh(\cos x)~dx=\int\sum\limits_{n=1}^\infty\dfrac{B_{2n}4^n(4^n-1)\cos^{2n-1}x}{(2n)!}dx$
For $\int\cos^{2n-1}x~dx$ , where $n$ is any natural number,
$\int\cos^{2n-1}x~dx$
$=\int\cos^{2n-2}x~d(\sin x)$
$=\int(1-\sin^2x)^{n-1}~d(\sin x)$
$=\int\sum\limits_{k=0}^{n-1}C_k^{n-1}(-1)^k\sin^{2k}x~d(\sin x)$
$=\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k(n-1)!\sin^{2k+1}x}{k!(n-k-1)!(2k+1)}+C$
$\therefore\int\sum\limits_{n=1}^\infty\dfrac{B_{2n}4^n(4^n-1)\cos^{2n-1}x}{(2n)!}dx=\sum\limits_{n=1}^\infty\sum\limits_{k=0}^{n-1}\dfrac{(-1)^kB_{2n}4^n(4^n-1)(n-1)!\sin^{2k+1}x}{(2n)!k!(n-k-1)!(2k+1)}+C$
Or you can directly use the formula http://upload.wikimedia.org/wikipedia/en/math/7/2/a/72a1058ad2087aec467af24bddcf9479.png:
$\int\tanh(\cos x)~dx=\int_0^x\tanh(\cos x)~dx+C=x\sum\limits_{n=1}^\infty\sum\limits_{k=1}^{2^n-1}\dfrac{(-1)^{k+1}}{2^n}\tanh\left(\cos\dfrac{kx}{2^n}\right)+C$
