## jennychan12 2 years ago integral from 0 to 1 of [e^x(cos(e^x))]dx ?

1. jennychan12

$\int\limits_{0}^{1} [e^x \cos (e^x)]dx$

2. KingGeorge

Do a u-substitution with $$u=\sin(e^x)$$, and your solution should pop right out.

3. hartnn

did you mean u=e^x ?

4. jennychan12

but if you do that, then du = e^xcos(e^x) and there's no cos (e^x) in the question.

5. hartnn

try u=e^x

6. jennychan12

that'd just be ucosu

7. KingGeorge

$$u=\sin(e^x)\implies du=e^x\cos(e^x)dx$$. So your integral becomes$\int du=u$

8. KingGeorge

Substitute back for $$u$$, and you get $\int_0^1 e^x\cos(e^x)dx=\sin(e^x)|_0^1$

9. hartnn

u =e^x du = e^x dx $$\int cos udu$$ ohh..now i could say that u =sin e^x is a better substitution..

10. jennychan12

oh wait. my bad. i thought u said u = sin u ok i see now

11. KingGeorge

Yup, I like to call my method "guessing the solution and proving you're right"

12. KingGeorge

Although if you really had no clue of the solution, $$u=e^x$$ would be a fine substitution. You would just have to integrate by parts.

13. hartnn

whenever i see a non-standard angle with sin/cos/.. i'll call that as 'bad angle' and put u= bad angle.... example : sin x^2 , cos log x ....

14. hartnn

how do we need integration by parts ?? O.o u= e^x du=e^xdx

15. KingGeorge

Oh. Right. You don't. Ignore my ramblings.