Integrate $\sin(\cos(x))$ with respect to $x$

Question

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 !!

using specific range of integration $-\pi,pi$ this looks suspicious like a besselfunction order 0 — tired, Oct 31 '15 at 15:57
@Dr.SonnhardGraubner is it not possible to get any approximation of this function ? — Pankaj Sejwal, Oct 31 '15 at 15:59
It's not invalid to integrate, it just doesn't have an elementary antiderivative. This is typical and still can be approximated. — Matt Samuel, Oct 31 '15 at 16:12
@MattSamuel Also there is no good documentation available anywhere. If you have some links to such approx functions, it could help. — Pankaj Sejwal, Oct 31 '15 at 16:24

Jack D'Aurizio · Accepted Answer · 2015-10-31T22:24:53.900

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:

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

Michael Medvinsky · Answer 2 · 2015-10-31T16:52:36.350

1

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 $\sin(\cos x)$.

edited Oct 31 '15 at 16:52

answered Oct 31 '15 at 16:46

Michael Medvinsky

5,798

2

I recommend $\cos x - \frac{\cos^3 x}{6}$, It's a lot closer. – Dylan Oct 31 '15 at 22:39

score 1 · Answer 3 · answered Nov 13 '15 at 16:13

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

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\cos^{2n+1}x}{(2n+1)!}dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\cos^{2n}x}{(2n+1)!}d(\sin x)$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(1-\sin^2x)^n}{(2n+1)!}d(\sin x)$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^n(-1)^k\sin^{2k}x}{(2n+1)!}d(\sin x)$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!\sin^{2k}x}{(2n+1)!k!(n-k)!}d(\sin x)$

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

Integrate $\sin(\cos(x))$ with respect to $x$

3 Answers3