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Use parts and solve for the unknown integral.

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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?

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Other answers:

Oh sorry
u = e^-x , dv = cos x
just so happens for this problem, either way works but in general, set u to non-trig and non-log functions
I got stuck at this part:|dw:1327222750455:dw|
do integration by parts again with integral of e^-x *sinx
|dw:1327223148495:dw| be careful of double negatives
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
\[e^(\ln \cos x-x)\]
this is only three steps
how is that ^^ easy to integrate ??
I need to use IBP for the assignment :-( @dumbcow: yes... but what happens to the e^(-x)? do I integrate that...?
lol cox anything raised to power of e , is that u take differential of anything in power wrt to x and divide on e
\[(cosx/-sinx-cosx)e^(\ln cosx-x)\]
like integral of e^-x is differentiate what ever is in power and divide by e so it comes out to b -e^-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..
@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
denebel, were you able to follow my work
Sorry, no..
|dw:1327225190587:dw| this is where i repeated integration by parts
\[let I=\int\limits_{}^{}e ^{-x}\cos x, also let u=e ^{-x} , du=-e ^{-x} , dv=\cos xdx, v=\sin x\]
\[so that I=e ^{-x}\sin x+\int\limits_{}^{}e ^{-x}\sin x\]
\[again let u=e ^{-x},du=e ^{-x}, dv=\sin x,v=-\cos xdx\]
\[\therefore I=e ^{-x}\sin x +[-e ^{-x}\cos x-\int\limits_{}^{}(e ^{-x}\cos xdx)\]] \[I=e ^{-x}\sin x+[-e ^{x}\cos x -I]\]
\[2I=e ^{-x}\sin x-e ^{-x}\cos x=e ^{-x}(\sin x-\cos x)\] \[I=[e ^{-x}(\sin x -\cos x)/2]\]

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