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AravindG Group TitleBest ResponseYou've already chosen the best response.1
\[ \large \int\limits e^x(xcos x+\sin x)dx\]
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
I have been stuck on this question for sometime..I am missing something somewhere.
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
@ganeshie8 , @jim_thompson5910 Can you help?
 one year ago

divu.mkr Group TitleBest ResponseYou've already chosen the best response.0
its a case in integration of by parts!!
 one year ago

surjithayer Group TitleBest ResponseYou've already chosen the best response.0
\[=\int\limits (x \cos x) e ^{x} dx+\int\limits (\sin x) e ^{x}dx\] \[=x \cos x e ^{x}\int\limits \left\{ x \left( \sin x \right)+\cos x *1 \right\}e ^{x} dx+\[\sin x e ^{x}\int\limits \cos x e ^{x}dx\]+c\] there is some mistake,try to find out.
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
I am thinking of doing it in the form e^x(f(x)+f'(x)) By supplying terms and subtracting.
 one year ago

surjithayer Group TitleBest ResponseYou've already chosen the best response.0
i have tried that also.
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
@surjithayer How did you get integral (xcos x)e^x dx=(xcos x)(e^x)
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
I am confused :/
 one year ago

divu.mkr Group TitleBest ResponseYou've already chosen the best response.0
you are doing that right and its integration is e^x.f(x)+c
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
@divu.mkr can you explain?
 one year ago

Euler271 Group TitleBest ResponseYou've already chosen the best response.1
everything i wrote is gone
 one year ago

Euler271 Group TitleBest ResponseYou've already chosen the best response.1
it all ends up being in terms of:\[\int\limits_{}^{}e^x cosx dx= \frac{ 1 }{ 2 }e^x(sinx + cosx) + C\] and\[\int\limits_{}^{}e^x sinx dx= \frac{ 1 }{ 2 }e^x(sinx  cosx) + C\]. a rather simple integration by parts. and seperating it like surji suggested and integrating only the first term by parts one gives:\[\frac{ x }{ 2 }e^x(sinx+cosx)  \frac{ 1 }{ 2 }\int\limits_{}^{} e^x (sinx + cosx)dx + C + \int\limits_{}^{}e^x sinx dx\]
 one year ago

Euler271 Group TitleBest ResponseYou've already chosen the best response.1
was my own fault. i pressed a wrong key and then backspace
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
Isnt it e^x (xcos x)?
 one year ago

Euler271 Group TitleBest ResponseYou've already chosen the best response.1
the integral of that, by parts once, gives the first two terms of the last line. taking u = x dv = e^xcosx
 one year ago

Euler271 Group TitleBest ResponseYou've already chosen the best response.1
i wouldn't mind solving it entirely if you're not convinced :)
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
yeah please :P
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
Its just I want to to do this qn in form e^x((fx)+f'(x)) Format and not apply parts.
 one year ago

Euler271 Group TitleBest ResponseYou've already chosen the best response.1
i'm not familiar with that method at all. only parts
 one year ago

Euler271 Group TitleBest ResponseYou've already chosen the best response.1
did you want to see parts or you know how?
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
yeah I want to see parts.
 one year ago

Euler271 Group TitleBest ResponseYou've already chosen the best response.1
i guess the only part is the:\[\int\limits_{}^{} e^x cosx dx\] u = e^x; du = e^x dx dv = cosx dx; v = sinx \[\int\limits_{}^{} e^x cosx dx = e^xsinx  \int\limits_{}^{} e^x sinx dx\] u = e^x; du = e^x dx dv = sinx dx; v = cosx \[\int\limits_{}^{}e^x cosx dx= e^x sinx  \left( e^x cosx + \int\limits_{}^{}e^x cosxdx \right)\] \[\int\limits_{}^{}e^x cosx dx = e^x(cosx + sinx)  \int\limits_{}^{}e^x cosx dx\]\[2\int\limits_{}^{}e^x cosx dx = e^x(sinx + cosx)\]\[\int\limits_{}^{}e^x cosx dx = \frac{ 1 }{ 2 }e^x (sinx + cosx)\]
 one year ago

divu.mkr Group TitleBest ResponseYou've already chosen the best response.0
do you have the answer..? i got one :D
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
yes final answer is 1/2e^x(xsin x+xcos xcos x)+c
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
@Euler271 Yeah I get it now thanks!
 one year ago

Euler271 Group TitleBest ResponseYou've already chosen the best response.1
glad i could help :)
 one year ago

surjithayer Group TitleBest ResponseYou've already chosen the best response.0
i have tried that also.
 one year ago

ybarrap Group TitleBest ResponseYou've already chosen the best response.0
Lots of integration by parts. Work on this in pieces (I've reused u and v several times but hopefully you can tell which is which by context): $$ \int e^x(x\cos x+\sin x)dx\\ \int xe^x \cos x ~dx+\int xe^x \sin x~dx\\ \text{For }\int xe^x \cos x ~dx\\ \text{Let }u=x,dv=e^x\cos x\\du=x,v=\int e^x\cos x dx\\ \text{For }\int e^x\cos x dx\\ \text{Let }u=e^x,dv=\cos x\\ du=e^x,v=\sin x\\ \text{Then }uv\int vdu=\\ e^x\sin x\int e^x\sin x dx\\ \int e^x\sin x dx=\\ \text{Let }u=e^x,dv=\sin x\\ \text{then }du=e^x,v=\cos x\\ uv\int vdu = \\ e^x\cos x+\int e^x\cos x dx\\ \text{Combining these results:}\\ \int e^x\cos x=\cfrac{e^x}{2}(\sin x+\cos x)\\ \text {So we now have }\\ \int xe^x \cos x ~dx\\ =\cfrac{xe^x}{2}(\sin x + \cos x)\int \cfrac{e^x}{2}(\sin x + \cos x)dx+\\ \qquad \cfrac{e^x}{2}(\sin x  \cos x)\\ \text{Take, }\int\frac{e^x}{2}(\sin x + \cos x)dx\\ \int\frac{e^x}{2}\sin x + \int \cfrac{e^x}{2}\cos xdx\\ =\cfrac{e^x}{4}(\sin x cos x)+\cfrac{e^x}{4}(\sin x +cos x)\\ =\cfrac{e^x}{2}\sin x\\ \text{So, }\\ \int xe^x \cos x ~dx\\ =\cfrac{xe^x}{2}(\sin x + \cos x)\cfrac{e^x}{2}\sin x+\cfrac{e^x}{2}(\sin x  \cos x)\\ =\cfrac{e^x}{2}(x\sin x + (x1)\cos x)\\ $$ Whew!! That's it!
 one year ago
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