# Integrating $\cos (x+\sin (x))$

I tried to solve $$\int\cos(x+\sin(x))\,dx$$ but it seems to be way out of my league (tried u-substitution with $u=x+\sin(x)$ and couldn't find an answer). Also, no one on the Internet seems to have tried this before and Wolfram|Alpha and Symbolab aren't helping that much. If anyone can help me, I would apreciate it very much.

• Maybe it does not have an anti derivative in terms of elementary functions... – imranfat Jun 26 '16 at 18:04
• @imranfat How could you prove that? and does one need high-level calculus knowledge to do it? – Nada F. Jun 26 '16 at 18:11
• I concur with turrati. HOWEVER.......sometimes there are integrals that do have an anti derivative in terms of elementary functions and I have seen those ones...so do not put all of your trust into a computer algebra system... – imranfat Jun 26 '16 at 19:00

The Jacobi-Anger expansion gives

$$e^{ix\sin\theta}=\sum_{n\in\mathbb{Z}}J_n(x) e^{ni\theta}\tag{1}$$ hence: $$e^{i\sin\theta+i\theta}=\sum_{n\in\mathbb{Z}} J_n(1)\,e^{(n+1)i\theta}\tag{2}$$ and by considering the real part of both sides: $$\cos(\theta+\sin\theta) = \sum_{n\in\mathbb{Z}} J_n(1)\,\cos((n+1)\theta)\tag{3}$$ so: $$\int\cos(\theta+\sin\theta)\,d\theta = \color{red}{C+\sum_{n\in\mathbb{Z}} \frac{J_{n}(1)}{n+1}\,\sin((n+1)\theta)}\tag{4}$$ where the RHS is a fast-converging series, since $J_n(1)$ behaves like $\frac{1}{2^n n!}$.
So, if you consider Bessel functions as elementary functions, there is a primitive in terms of elementary functions.

• Does Σ_{n ∈ Z} mean that the sum runs through all the integers (positives and negatives)? – Nada F. Jun 26 '16 at 20:25
• @MassimilianoTron: yes. That can be simplified since $J_{-n}(z)=(-1)^n J_{n}(z)$. – Jack D'Aurizio Jun 26 '16 at 20:27

If you look in wolframalpha it says that this integral can not be expressed in elementary math functions, so the only way to express is the integral itself.

This is the wolfram result: https://www.wolframalpha.com/input/?i=integral+cos(x%2Bsin(x))