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

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Denebel
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1327221848252:dw I have trouble determining which is u?

Aron_West
 3 years ago
Best ResponseYou've already chosen the best response.0\[1/2 [e^{x} (sinxcosx)]+constant \]

Denebel
 3 years ago
Best ResponseYou've already chosen the best response.0Lol. I need to do IBP. which is the u?

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.1just so happens for this problem, either way works but in general, set u to nontrig and nonlog functions

Denebel
 3 years ago
Best ResponseYou've already chosen the best response.0I got stuck at this part:dw:1327222750455:dw

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.1do integration by parts again with integral of e^x *sinx

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1327223148495:dw be careful of double negatives

wasiqss
 3 years ago
Best ResponseYou've already chosen the best response.0i solved it b some diff method take ln both sides, comes out to b lny=lncosxx and it also equals to e^lncosxx=y hence now its very easy to integrate

wasiqss
 3 years ago
Best ResponseYou've already chosen the best response.0this is only three steps

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.1how is that ^^ easy to integrate ??

Denebel
 3 years ago
Best ResponseYou've already chosen the best response.0I need to use IBP for the assignment :( @dumbcow: yes... but what happens to the e^(x)? do I integrate that...?

wasiqss
 3 years ago
Best ResponseYou've already chosen the best response.0lol cox anything raised to power of e , is that u take differential of anything in power wrt to x and divide on e

wasiqss
 3 years ago
Best ResponseYou've already chosen the best response.0\[(cosx/sinxcosx)e^(\ln cosxx)\]

wasiqss
 3 years ago
Best ResponseYou've already chosen the best response.0like integral of e^x is differentiate what ever is in power and divide by e so it comes out to b e^x

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.1repeat integration by parts u = e^x , dv = sin x du = e^x, v = cos(x) dw:1327223482801:dw

wasiqss
 3 years ago
Best ResponseYou've already chosen the best response.0dumbcow in mine only one integration and fairly easy one

Denebel
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, wait, there isn't a double negative right? Because dw:1327223660348:dw

Denebel
 3 years ago
Best ResponseYou've already chosen the best response.0Wait I am going backwards..

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.1@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
 3 years ago
Best ResponseYou've already chosen the best response.1denebel, were you able to follow my work

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1327225190587:dw this is where i repeated integration by parts

mark_o.
 3 years ago
Best ResponseYou've already chosen the best response.0\[let I=\int\limits_{}^{}e ^{x}\cos x, also let u=e ^{x} , du=e ^{x} , dv=\cos xdx, v=\sin x\]

mark_o.
 3 years ago
Best ResponseYou've already chosen the best response.0\[so that I=e ^{x}\sin x+\int\limits_{}^{}e ^{x}\sin x\]

mark_o.
 3 years ago
Best ResponseYou've already chosen the best response.0\[again let u=e ^{x},du=e ^{x}, dv=\sin x,v=\cos xdx\]

mark_o.
 3 years ago
Best ResponseYou've already chosen the best response.0\[\therefore I=e ^{x}\sin x +[e ^{x}\cos x\int\limits_{}^{}(e ^{x}\cos xdx)\]] \[I=e ^{x}\sin x+[e ^{x}\cos x I]\]

mark_o.
 3 years ago
Best ResponseYou've already chosen the best response.0\[2I=e ^{x}\sin xe ^{x}\cos x=e ^{x}(\sin x\cos x)\] \[I=[e ^{x}(\sin x \cos x)/2]\]
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