Question: > 
> Evaluate $\int \cos(\cos x)~dx$
> 
> 
> 

I tried to use chain rule but failed. Can anyone help me please?

Answer: 
This is probably too long for a comment. 

Wolfram alpha indicates that the solution has the form
$$\sum\_{n=0}^{\infty} \frac{x^{2n+1}(a\_{n}\sin(1)+b\_{n}\cos(1))}{(2n+1)!}$$

The $-a\_{n}$ appear to correspond to [oeis:A192007](https://oeis.org/A192007), e.g.f.: $\sin(\cos(x)-1)$ (even part), and the $b\_{n}$ appear to correspond to [oeis:A192060](https://oeis.org/A192060). e.g.f: $\cos(\cos(1)-1)$ (even part)
$\int\cos\cos x~dx=\int\sum\limits\_{n=0}^\infty\dfrac{(-1)^n\cos^{2n}x}{(2n)!}~dx=\int\left(1+\sum\limits\_{n=1}^\infty\dfrac{(-1)^n\cos^{2n}x}{(2n)!}\right)~dx$

For $n$ 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.

$\therefore\int\left(1+\sum\limits\_{n=1}^\infty\dfrac{(-1)^n\cos^{2n}x}{(2n)!}\right)~dx$

$=x+\sum\limits\_{n=1}^\infty\dfrac{x}{4^n(n!)^2}+\sum\limits\_{n=1}^\infty\sum\limits\_{k=1}^n\dfrac{(-1)^n((k-1)!)^2\sin x\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$

$=\sum\limits\_{n=0}^\infty\dfrac{x}{4^n(n!)^2}+\sum\limits\_{n=1}^\infty\sum\limits\_{k=1}^n\dfrac{(-1)^n((k-1)!)^2\sin x\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$