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

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

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

Oh sorry

u = e^-x , dv = cos x

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

|dw:1327222868911:dw|

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

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

\[e^(\ln \cos x-x)\]

this is only three steps

how is that ^^ easy to integrate ??

\[(cosx/-sinx-cosx)e^(\ln cosx-x)\]

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

dumbcow in mine only one integration and fairly easy one

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

Wait I am going backwards..

denebel,
were you able to follow my work

Sorry, no..

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

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

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

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

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