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AravindG

  • 2 years ago

integral (e^x(xcos x+sin x))dx=?

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  1. AravindG
    • 2 years ago
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    \[ \large \int\limits e^x(xcos x+\sin x)dx\]

  2. AravindG
    • 2 years ago
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    I have been stuck on this question for sometime..I am missing something somewhere.

  3. AravindG
    • 2 years ago
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    @ganeshie8 , @jim_thompson5910 Can you help?

  4. divu.mkr
    • 2 years ago
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    its a case in integration of by parts!!

  5. surjithayer
    • 2 years ago
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    \[=\int\limits (x \cos x) e ^{x} dx+\int\limits (\sin x) e ^{x}dx\] \[=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.

  6. AravindG
    • 2 years ago
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    I am thinking of doing it in the form e^x(f(x)+f'(x)) By supplying terms and subtracting.

  7. surjithayer
    • 2 years ago
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    i have tried that also.

  8. AravindG
    • 2 years ago
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    @surjithayer How did you get integral (xcos x)e^x dx=(xcos x)(e^x)

  9. AravindG
    • 2 years ago
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    I am confused :/

  10. divu.mkr
    • 2 years ago
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    you are doing that right and its integration is e^x.f(x)+c

  11. AravindG
    • 2 years ago
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    @divu.mkr can you explain?

  12. Euler271
    • 2 years ago
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    everything i wrote is gone

  13. AravindG
    • 2 years ago
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    How??

  14. Euler271
    • 2 years ago
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    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:\[\frac{ x }{ 2 }e^x(sinx+cosx) - \frac{ 1 }{ 2 }\int\limits_{}^{} e^x (sinx + cosx)dx + C + \int\limits_{}^{}e^x sinx dx\]

  15. Euler271
    • 2 years ago
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    was my own fault. i pressed a wrong key and then backspace

  16. Euler271
    • 2 years ago
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    once**

  17. AravindG
    • 2 years ago
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    Isnt it e^x (xcos x)?

  18. Euler271
    • 2 years ago
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    the integral of that, by parts once, gives the first two terms of the last line. taking u = x dv = e^xcosx

  19. Euler271
    • 2 years ago
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    i wouldn't mind solving it entirely if you're not convinced :)

  20. AravindG
    • 2 years ago
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    yeah please :P

  21. AravindG
    • 2 years ago
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    Its just I want to to do this qn in form e^x((fx)+f'(x)) Format and not apply parts.

  22. Euler271
    • 2 years ago
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    i'm not familiar with that method at all. only parts

  23. Euler271
    • 2 years ago
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    did you want to see parts or you know how?

  24. AravindG
    • 2 years ago
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    yeah I want to see parts.

  25. Euler271
    • 2 years ago
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    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 \[\int\limits_{}^{}e^x cosx dx= e^x sinx - \left( -e^x cosx + \int\limits_{}^{}e^x cosxdx \right)\] \[\int\limits_{}^{}e^x cosx dx = e^x(cosx + sinx) - \int\limits_{}^{}e^x cosx dx\]\[2\int\limits_{}^{}e^x cosx dx = e^x(sinx + cosx)\]\[\int\limits_{}^{}e^x cosx dx = \frac{ 1 }{ 2 }e^x (sinx + cosx)\]

  26. divu.mkr
    • 2 years ago
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    do you have the answer..? i got one :D

  27. AravindG
    • 2 years ago
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    yes final answer is 1/2e^x(xsin x+xcos x-cos x)+c

  28. AravindG
    • 2 years ago
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    @Euler271 Yeah I get it now thanks!

  29. Euler271
    • 2 years ago
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    glad i could help :)

  30. surjithayer
    • 2 years ago
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    i have tried that also.

  31. ybarrap
    • 2 years ago
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    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): $$ \int e^x(x\cos x+\sin x)dx\\ \int xe^x \cos x ~dx+\int xe^x \sin x~dx\\ \text{For }\int xe^x \cos x ~dx\\ \text{Let }u=x,dv=e^x\cos x\\du=x,v=\int e^x\cos x dx\\ \text{For }\int e^x\cos x dx\\ \text{Let }u=e^x,dv=\cos x\\ du=e^x,v=\sin x\\ \text{Then }uv-\int vdu=\\ e^x\sin x-\int e^x\sin x dx\\ \int e^x\sin x dx=\\ \text{Let }u=e^x,dv=\sin x\\ \text{then }du=e^x,v=-\cos x\\ uv-\int vdu = \\ -e^x\cos x+\int e^x\cos x dx\\ \text{Combining these results:}\\ \int e^x\cos x=\cfrac{e^x}{2}(\sin x+\cos x)\\ \text {So we now have }\\ \int xe^x \cos x ~dx\\ =\cfrac{xe^x}{2}(\sin x + \cos x)-\int \cfrac{e^x}{2}(\sin x + \cos x)dx+\\ \qquad \cfrac{e^x}{2}(\sin x - \cos x)\\ \text{Take, }\int\frac{e^x}{2}(\sin x + \cos x)dx\\ \int\frac{e^x}{2}\sin x + \int \cfrac{e^x}{2}\cos xdx\\ =\cfrac{e^x}{4}(\sin x -cos x)+\cfrac{e^x}{4}(\sin x +cos x)\\ =\cfrac{e^x}{2}\sin x\\ \text{So, }\\ \int xe^x \cos x ~dx\\ =\cfrac{xe^x}{2}(\sin x + \cos x)-\cfrac{e^x}{2}\sin x+\cfrac{e^x}{2}(\sin x - \cos x)\\ =\cfrac{e^x}{2}(x\sin x + (x-1)\cos x)\\ $$ Whew!! That's it!

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