## Denebel 3 years ago Use parts and solve for the unknown integral.

1. Denebel

|dw:1327221848252:dw| I have trouble determining which is u?

2. Aron_West

$1/2 [e^{-x} (sinx-cosx)]+constant$

3. Denebel

Lol. I need to do IBP. which is the u?

4. Aron_West

Oh sorry

5. dumbcow

u = e^-x , dv = cos x

6. dumbcow

just so happens for this problem, either way works but in general, set u to non-trig and non-log functions

7. Denebel

I got stuck at this part:|dw:1327222750455:dw|

8. Denebel

|dw:1327222868911:dw|

9. dumbcow

do integration by parts again with integral of e^-x *sinx

10. dumbcow

|dw:1327223148495:dw| be careful of double negatives

11. wasiqss

i solved it b some diff method take ln both sides, comes out to b lny=lncosx-x and it also equals to e^lncosx-x=y hence now its very easy to integrate

12. wasiqss

$e^(\ln \cos x-x)$

13. wasiqss

this is only three steps

14. dumbcow

how is that ^^ easy to integrate ??

15. Denebel

I need to use IBP for the assignment :-( @dumbcow: yes... but what happens to the e^(-x)? do I integrate that...?

16. wasiqss

lol cox anything raised to power of e , is that u take differential of anything in power wrt to x and divide on e

17. wasiqss

$(cosx/-sinx-cosx)e^(\ln cosx-x)$

18. wasiqss

like integral of e^-x is differentiate what ever is in power and divide by e so it comes out to b -e^-x

19. dumbcow

repeat integration by parts u = e^-x , dv = sin x du = -e^-x, v = -cos(x) |dw:1327223482801:dw|

20. wasiqss

dumbcow in mine only one integration and fairly easy one

21. Denebel

Oh, wait, there isn't a double negative right? Because |dw:1327223660348:dw|

22. Denebel

Wait I am going backwards..

23. dumbcow

@wasiqss: it doesn't work like that when integrating if the derivative of the exponent is not a constant Example: integral e^(x^2) dx != e^(x^2)/2x "!=" means not equal

24. dumbcow

denebel, were you able to follow my work

25. Denebel

Sorry, no..

26. dumbcow

|dw:1327225190587:dw| this is where i repeated integration by parts

27. mark_o.

$let I=\int\limits_{}^{}e ^{-x}\cos x, also let u=e ^{-x} , du=-e ^{-x} , dv=\cos xdx, v=\sin x$

28. mark_o.

$so that I=e ^{-x}\sin x+\int\limits_{}^{}e ^{-x}\sin x$

29. mark_o.

$again let u=e ^{-x},du=e ^{-x}, dv=\sin x,v=-\cos xdx$

30. mark_o.

$\therefore I=e ^{-x}\sin x +[-e ^{-x}\cos x-\int\limits_{}^{}(e ^{-x}\cos xdx)$] $I=e ^{-x}\sin x+[-e ^{x}\cos x -I]$

31. mark_o.

$2I=e ^{-x}\sin x-e ^{-x}\cos x=e ^{-x}(\sin x-\cos x)$ $I=[e ^{-x}(\sin x -\cos x)/2]$