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

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

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

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

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

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

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

wasiqssBest ResponseYou've already chosen the best response.0
i 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
 2 years ago

wasiqssBest ResponseYou've already chosen the best response.0
this is only three steps
 2 years ago

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

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

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

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

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

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

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

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

DenebelBest ResponseYou've already chosen the best response.0
Wait I am going backwards..
 2 years ago

dumbcowBest 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
 2 years ago

dumbcowBest ResponseYou've already chosen the best response.1
denebel, were you able to follow my work
 2 years ago

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

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

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

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

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

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