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

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

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

surjithayer Group TitleBest ResponseYou've already chosen the best response.0
i have tried that also.
 11 months 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)
 11 months ago

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

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

Euler271 Group TitleBest ResponseYou've already chosen the best response.1
everything i wrote is gone
 11 months 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:\[\frac{ x }{ 2 }e^x(sinx+cosx)  \frac{ 1 }{ 2 }\int\limits_{}^{} e^x (sinx + cosx)dx + C + \int\limits_{}^{}e^x sinx dx\]
 11 months ago

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

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

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

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

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

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

AravindG Group TitleBest ResponseYou've already chosen the best response.1
yeah I want to see parts.
 11 months 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 \[\int\limits_{}^{}e^x cosx dx= e^x sinx  \left( e^x cosx + \int\limits_{}^{}e^x cosxdx \right)\] \[\int\limits_{}^{}e^x cosx dx = e^x(cosx + sinx)  \int\limits_{}^{}e^x cosx dx\]\[2\int\limits_{}^{}e^x cosx dx = e^x(sinx + cosx)\]\[\int\limits_{}^{}e^x cosx dx = \frac{ 1 }{ 2 }e^x (sinx + cosx)\]
 11 months ago

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

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

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

surjithayer Group TitleBest ResponseYou've already chosen the best response.0
i have tried that also.
 11 months 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!
 11 months ago
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