# Integrate cos(x^5)

1. Oct 19, 2007

### Artaxerxes

Can you solve this $\int \cos x^5 dx$ ?

2. Oct 19, 2007

### arildno

No, I can't

3. Oct 19, 2007

### rocomath

nope ...

4. Oct 19, 2007

### Zurtex

It probably can't be done in terms of elementary functions. Mathematica has an answer of 2 different forms:

Take E(v,z) here to be this function http://functions.wolfram.com/GammaBetaErf/ExpIntegralE/ we have:

$$\int \cos x^5 dx = - \frac{1}{10} \left( E\left( \frac{4}{5} , -i x^5 \right) + E\left( \frac{4}{5} , i x^5 \right) \right)$$

Or in terms of the gamma function http://functions.wolfram.com/GammaBetaErf/Gamma/ :

$$\int \cos x^5 dx = - \frac{\left( x^{10} \right)^{\frac{4}{5}}}{10x^9}\left( \left(i x^5 \right)^{\frac{1}{5}} \Gamma \left( \frac{1}{5} , -i x^5 \right) + \left(-i x^5 \right)^{\frac{1}{5}} \Gamma \left( \frac{1}{5} , i x^5 \right) \right)$$

I don't really understand why mathematica didn't simplify it further, so I've tried to keep it to what mathematica outputted.

Last edited: Oct 19, 2007
5. Oct 19, 2007

### Artaxerxes

Thank you!

6. Oct 19, 2007

### DavidSmith

thats why series solutions are so conviennt

7. Oct 20, 2007

### Gib Z

The Taylor series for cos converges for all real x, so you should have just let x^5 = u in the taylor expansion of cos u, integrated term by term and you are left with an even nicer result than what mathematicia gives out.