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AravindG
 one year ago
integral (e^x(xcos x+sin x))dx=?
AravindG
 one year ago
integral (e^x(xcos x+sin x))dx=?

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

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

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

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

surjithayer
 one year ago
Best 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.

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

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

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

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

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

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

Euler271
 one year ago
Best ResponseYou've already chosen the best response.1it 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:\[\frac{ x }{ 2 }e^x(sinx+cosx)  \frac{ 1 }{ 2 }\int\limits_{}^{} e^x (sinx + cosx)dx + C + \int\limits_{}^{}e^x sinx dx\]

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

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

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

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

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

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

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

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

Euler271
 one year ago
Best ResponseYou've already chosen the best response.1i 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 \[\int\limits_{}^{}e^x cosx dx= e^x sinx  \left( e^x cosx + \int\limits_{}^{}e^x cosxdx \right)\] \[\int\limits_{}^{}e^x cosx dx = e^x(cosx + sinx)  \int\limits_{}^{}e^x cosx dx\]\[2\int\limits_{}^{}e^x cosx dx = e^x(sinx + cosx)\]\[\int\limits_{}^{}e^x cosx dx = \frac{ 1 }{ 2 }e^x (sinx + cosx)\]

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

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

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

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

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0Lots 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!
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