Integral of cos(e^x)

1. Oct 4, 2007

rocomath

$$\int\cos{e^x}dx$$

let

$$u=\cos{e^x}$$
$$du=-e^{x}\sin{e^x}dx$$

$$dv=dx$$
$$v=x$$

$$\int\cos{e^x}dx=x\cos{e^x}+\int xe^{x}\sin{e^x}dx$$

Last edited: Oct 4, 2007
2. Oct 4, 2007

Hurkyl

Staff Emeritus
Well, that's generally what happens when you do the same integration by parts first forwards and then backwards: you wind up right where you started.

3. Oct 5, 2007

Avodyne

This does not have a closed-form solution in terms of simple functions. Are you sure you copied it correctly?

4. Oct 5, 2007

rocomath

yep, i'm in Calc 2, so i prob wouldn't be able to solve it even with a little guidance?

5. Oct 5, 2007

Avodyne

Well, there's not much to do. Make the substitution x=ln(u), dx=du/u, and you get cos(u)/u. The integral of this is the Cosine Integral function:
http://mathworld.wolfram.com/CosineIntegral.html

6. Oct 5, 2007

SanjeevGupta

What about expanding the cosine out and then integrating the infinite series. At least that gives you a result, however ugly.

7. Oct 5, 2007

rocomath

i was thinking of using this approach but i wasn't sure if i could do that

i also tried that website after class and i was like huh :D but thanks.

8. Oct 5, 2007

rocomath

i'm not on that section yet so i'll have to wait.

9. Oct 5, 2007

real10

ar u sure it's not $$cos(x)*e^x$$ ?? for Calc 2 cos(e^x) seems kindof hard given that u havent done series yet...