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$$\int \tanh \left ( \cos x \right )\mathrm{d}x$$

Is this analytically Solvable?

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This is not wolframalpha :) –  Dominic Michaelis Jul 15 '13 at 8:03
    
And Wolframalpha says no –  Dominic Michaelis Jul 15 '13 at 8:07
    
How did you land on this integral? Also, I've found integral-table.com/integral-table.html to be fairly useful. –  dfeuer Jul 15 '13 at 8:39

1 Answer 1

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:

$\int\tanh(\cos x)~dx=\int_0^x\tanh(\cos x)~dx+C=x\sum\limits_{n=1}^\infty\sum\limits_{k=1}^{2^n-1}\dfrac{(-1)^{k+1}}{2^n}\tanh\left(\cos\dfrac{kx}{2^n}\right)+C$

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