Integrate $\sin(\cos(x))$ with respect to $x$ Solve this:

$$\int\sin(\cos x)dx$$

I checked on Maxima, mathematica but both cannot find its integral though numerical approximation is available in later. Has someone faced similar problem? Also, I saw that such problems are not stated in books also, no function can be expected to be invalid to integrate !!
 A: Bessel functions arise exactly from the Fourier transform of inverse trigonometric functions. For instance, $\sin(\cos(x))$ is a $2\pi$-periodic even function, and it can be expanded as a Fourier cosine series:
$$ \sin(\cos(x)) = 2\sum_{n\geq 0} (-1)^n J_{2n+1}(1)\,\cos\left((2n+1) x\right)\tag{1} $$
where $J_{2n+1}(1)$, given by the value of a Bessel function of the first kind, decays very fast as $n$ approaches $+\infty$:
$$ J_n(1) \leq \frac{1}{2^n n!}.\tag{2} $$
$(1)$ gives that an anti-derivative of $\sin(\cos(x))$ is given by the extremely fast convergent series:

$$ \int \sin(\cos(x))\,dx = \sum_{n\geq 0}\frac{(-1)^n J_{2n+1}(1)}{n+\frac{1}{2}}\,\sin((2n+1)x).\tag{3}$$

A: Reffering to question in comments.
In general, In order to numerically approximate integral of given function one approximate the function using polynomials and then integrate the polynomial. 
You don't need however to start from zero, depends on how good you want to approximate you can go to midpoint, trapezoidal or Simpson integration rules or even Gauss quadrature. A book\course in numerical analysis could be useful.
If you do not run after preciseness you can approximate it by integrating $$\sin(1)\cos(x)$$
which is visually very close to $\sin(\cos x)$. 
A: $\int\sin(\cos x)~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\cos^{2n+1}x}{(2n+1)!}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\cos^{2n}x}{(2n+1)!}d(\sin x)$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(1-\sin^2x)^n}{(2n+1)!}d(\sin x)$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^n(-1)^k\sin^{2k}x}{(2n+1)!}d(\sin x)$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!\sin^{2k}x}{(2n+1)!k!(n-k)!}d(\sin x)$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!\sin^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$
