Evaluate $\int \cos(\cos x)~dx$
I tried to use chain rule but failed. Can anyone help me please?
$\endgroup$ 24 Answers
$\begingroup$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, e.g.f.: $\sin(\cos(x)-1)$ (even part), and the $b_{n}$ appear to correspond to oeis:A192060. e.g.f: $\cos(\cos(1)-1)$ (even part)
$\endgroup$ $\begingroup$The indefinite integral has no simpler form (known), but there are some definite integrals, like this $$ \int_0^{\pi/2} \cos(\cos x)\,dx = \frac{\pi}{2}\;J_0(1) $$ in terms of a Bessel function.
$\endgroup$ $\begingroup$This integral doesn't have a nice closed-form solution in terms of elementary functions, so this question is impossible (assuming you're just supposed to find the antiderivative in a form simpler than $\int \cos(\cos(x)) dx$)
$\endgroup$ $\begingroup$$\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.
$\therefore\int\left(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\cos^{2n}x}{(2n)!}\right)~dx$
$=x+\sum\limits_{n=1}^\infty\dfrac{x}{4^n(n!)^2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n((k-1)!)^2\sin x\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$
$=\sum\limits_{n=0}^\infty\dfrac{x}{4^n(n!)^2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n((k-1)!)^2\sin x\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$
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