Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
1  
Meta: Both (integral) and (integration) tags are used here. Shouldn't only one tag suffice? But I'm not sure which. –  user2468 Mar 7 '12 at 14:31
1  
Not sure if this is relevant. Mathematica documentation claims that $\int \sin(\sin x) d x$ "can in principle be represented as an infinite sum of $_{2}F_{1}$ hypergeometric functions, or as a suitably generalized Kampé de Fériet hypergeometric function of two variables." –  sdcvvc Mar 7 '12 at 17:07
    
@jasoncube You have to accept some of the answers of your questions . Otherwise fewer people will be interested in answering you. You acceptance rate is 0%. –  Kirthi Raman Apr 23 '12 at 11:11
add comment

3 Answers

up vote 12 down vote accepted

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)

share|improve this answer
add comment

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.

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.