Denebel
Use parts and solve for the unknown integral.
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Denebel
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|dw:1327221848252:dw|
I have trouble determining which is u?
Aron_West
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\[1/2 [e^{-x} (sinx-cosx)]+constant \]
Denebel
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Lol. I need to do IBP. which is the u?
Aron_West
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Oh sorry
dumbcow
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u = e^-x , dv = cos x
dumbcow
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just so happens for this problem, either way works
but in general, set u to non-trig and non-log functions
Denebel
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I got stuck at this part:|dw:1327222750455:dw|
Denebel
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|dw:1327222868911:dw|
dumbcow
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do integration by parts again with integral of e^-x *sinx
dumbcow
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|dw:1327223148495:dw|
be careful of double negatives
wasiqss
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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
wasiqss
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\[e^(\ln \cos x-x)\]
wasiqss
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this is only three steps
dumbcow
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how is that ^^ easy to integrate ??
Denebel
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I need to use IBP for the assignment :-(
@dumbcow: yes... but what happens to the e^(-x)? do I integrate that...?
wasiqss
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lol cox anything raised to power of e , is that u take differential of anything in power wrt to x and divide on e
wasiqss
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\[(cosx/-sinx-cosx)e^(\ln cosx-x)\]
wasiqss
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like integral of e^-x is
differentiate what ever is in power and divide by e so it comes out to b
-e^-x
dumbcow
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repeat integration by parts
u = e^-x , dv = sin x
du = -e^-x, v = -cos(x)
|dw:1327223482801:dw|
wasiqss
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dumbcow in mine only one integration and fairly easy one
Denebel
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Oh, wait, there isn't a double negative right? Because |dw:1327223660348:dw|
Denebel
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Wait I am going backwards..
dumbcow
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@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
dumbcow
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denebel,
were you able to follow my work
Denebel
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Sorry, no..
dumbcow
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|dw:1327225190587:dw|
this is where i repeated integration by parts
mark_o.
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\[let I=\int\limits_{}^{}e ^{-x}\cos x, also let u=e ^{-x} , du=-e ^{-x} , dv=\cos xdx, v=\sin x\]
mark_o.
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\[so that I=e ^{-x}\sin x+\int\limits_{}^{}e ^{-x}\sin x\]
mark_o.
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\[again let u=e ^{-x},du=e ^{-x}, dv=\sin x,v=-\cos xdx\]
mark_o.
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\[\therefore I=e ^{-x}\sin x +[-e ^{-x}\cos x-\int\limits_{}^{}(e ^{-x}\cos xdx)\]]
\[I=e ^{-x}\sin x+[-e ^{x}\cos x -I]\]
mark_o.
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\[2I=e ^{-x}\sin x-e ^{-x}\cos x=e ^{-x}(\sin x-\cos x)\]
\[I=[e ^{-x}(\sin x -\cos x)/2]\]