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Can't figure out this integral

  1. Feb 17, 2012 #1
    I am unsure how to solve this integral...

    [itex]\int5e^{\frac{t}{2}}sin(2t) dt[/itex]


    Work I did:

    [itex]5\int e^{\frac{t}{2}}sin(2t) dt[/itex]

    integration by parts wont work I don't think therefore I am stuck.
     
  2. jcsd
  3. Feb 17, 2012 #2
    Re: Cant figure out this integral

    Try to let u = e^(t/2) and dv = sin2tdt, then the resulting integral will have a tail Integral (1/2cost(1/2e^(t/2)dt). Use the same process and eventually it will go back to the same integral on the problem. Then just use algebraic technique to solve the problem. Hope it helps. (im still at school at the moment, i'll figure this out when i reach home.)
     
  4. Feb 17, 2012 #3
    Re: Cant figure out this integral

    5∫et/2sin(2t)dt

    Let:
    u = sin(2t), therefore du = 2cos(2t)
    vdv = et/2 and v = 2et/2

    Hope this helps!
     
    Last edited by a moderator: Feb 17, 2012
  5. Feb 17, 2012 #4
    Re: Cant figure out this integral

    Was able to get to:

    [itex]5 \int e^{\frac{t}{2}}sin(2t) = 2sin(2t)e^{\frac{t}{2}}-4\int e^{\frac{t}{2}}cos(2t)[/itex]
    [itex]5 \int e^{\frac{t}{2}}sin(2t) = 2sin(2t)e^{\frac{t}{2}}-4[2cos(2t)e^{\frac{t}{2}} + 4\int e^{\frac{t}{2}}sin(2t) ][/itex]
    not sure what to do now.
     
    Last edited: Feb 17, 2012
  6. Feb 17, 2012 #5
    Re: Cant figure out this integral

    I haven't checked your wwork but at this point you just need to rearrange the equation so your integral only appears on the left, you're so close
    :-)
     
  7. Feb 17, 2012 #6
    Re: Cant figure out this integral

    [itex] u = sin(2t)[/itex]
    [itex] du = 2cos(2t)[/itex]
    [itex] v = 2e^{\frac{t}{2}}[/itex]
    [itex] dv = e^{\frac{t}{2}}[/itex]


    [itex]5 \int e^{\frac{t}{2}}sin(2t) = 2sin(2t)e^{\frac{t}{2}}-4\int e^{\frac{t}{2}}cos(2t)[/itex]

    [itex]u = cos(2t)[/itex]
    [itex]du = -2sin(2t)[/itex]

    [itex]v=2e^{\frac{t}{2}}[/itex]
    [itex]dv=e^{\frac{t}{2}}[/itex]

    [itex]5 \int e^{\frac{t}{2}}sin(2t) = 2sin(2t)e^{\frac{t}{2}}-4[-2cos(2t)e^{\frac{t}{2}} + 4\int e^{\frac{t}{2}}sin(2t) ][/itex]

    [itex]5 \int e^{\frac{t}{2}}sin(2t) = 2sin(2t)e^{\frac{t}{2}}+8cos(2t)e^{\frac{t}{2}} -16\int e^{\frac{t}{2}}sin(2t) [/itex]

    What do I do with the:
    [itex]-16\int e^{\frac{t}{2}}sin(2t)[/itex]
     
  8. Feb 17, 2012 #7
    Re: Cant figure out this integral

    Move it to the left side, it might be easier to replace it with a 'j' or some such since all you have left to do is some algebra and you don't get distracted by the integral sign
     
  9. Feb 17, 2012 #8
    Re: Cant figure out this integral

    moving it over by adding to both sides I get :
    [itex]21 \int e^{\frac{t}{2}}sin(2t) = 2sin(2t)e^{\frac{t}{2}}+8cos(2t)e^{\frac{t}{2}}[/itex]

    this doesn't look right if i divide by 21 after this? I'm doing something wrong here I think.
     
    Last edited: Feb 17, 2012
  10. Feb 17, 2012 #9

    vela

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    Re: Cant figure out this integral

    That's what you want to do, but you forgot the overall factor of 5 on the righthand side when you integrated by parts. So fix that first.
     
  11. Feb 17, 2012 #10
    Re: Cant figure out this integral

    I don't know how, or I would of already done it.
     
  12. Feb 17, 2012 #11

    vela

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    You don't know how to multiply the righthand side by 5?
     
  13. Feb 17, 2012 #12
    Why would you multiply by 5 instead of dividing by 5? To move stuff don't you do the inverse.
     
  14. Feb 17, 2012 #13

    vela

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    I'm talking about the very beginning of your work.

    What is ##2 e^{t/2}\sin 2t - 4 \int e^{t/2}\cos 2t\,dt##, which is what you got after the first integration by parts, equal to? Is it
    $$\int e^{t/2}\sin 2t\, dt$$ or $$5 \int e^{t/2}\sin 2t\, dt?$$
     
  15. Feb 17, 2012 #14
    its equal to

    $$5 \int e^{t/2}\sin 2t\, dt$$
     
  16. Feb 17, 2012 #15

    vela

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    So tell me then, what is ##\int e^{t/2}\sin 2t\, dt## equal to after integrating by parts once?
     
  17. Feb 17, 2012 #16
    its equal to

    [itex] uv - \int v du [/itex]

    where
    [itex]u = sin(2t)[/itex]
    [itex]du = 2cos(2t)[/itex]
    [itex]v = 2e^{\frac{t}{2}}[/itex]
    [itex]dv = e^{\frac{t}{2}}[/itex]

    Giving

    [itex] 2sin(2t)e^{\frac{t}{2}} - 4\int e^{\frac{t}{2}}cos(2t)[/itex]
     
  18. Feb 17, 2012 #17

    vela

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    OK, in post 14, you said
    $$ 2 e^{t/2}\sin 2t - 4 \int e^{t/2}\cos 2t\,dt = 5 \int e^{t/2}\sin 2t\, dt.$$
    In post 14, you said
    $$ 2 e^{t/2}\sin 2t - 4 \int e^{t/2}\cos 2t\,dt = \int e^{t/2}\sin 2t\, dt.$$
    Since the lefthand sides are the same, you're saying
    $$ \int e^{t/2}\sin 2t\, dt = 5 \int e^{t/2}\sin 2t\, dt$$ Can you see the problem?
     
  19. Feb 17, 2012 #18
    just skimming over, I see it hasn't been suggested to use the exponential form of Sin(2t)

    Any time I see an integral over a Cos or a Sin I find that it takes me a lot less time just to use the fact that
    [itex]Sin(at) = \frac{e^{i\ a\ t} - e^{-i\ a\ t}}{2\ i}[/itex]
    Then you end up with two easy integrals of exopnents which you can then turn back into trig functions after everything has been done.

    That might just be me though, most other people I know prefer u substitution.
     
  20. Feb 17, 2012 #19
    from applying by parts once you can see "cos x" and "sin x" thus use by parts once again on integral part !
     
  21. Feb 17, 2012 #20
    I can see that the two sides
    $$ \int e^{t/2}\sin 2t\, dt = 5 \int e^{t/2}\sin 2t\, dt$$
    are not equal to each other.

    So, since i found the original equation within the series of doing integration by parts over and over. And its not equal to $$5 \int e^{t/2}\sin 2t\, dt$$ I have to multiply the entire side by 5 ?

    giving

    [itex]5 \int e^{\frac{t}{2}}sin(2t) = (5)2sin(2t)e^{\frac{t}{2}}+(5)8cos(2t)e^{\frac{t}{2}} -16(5)\int e^{\frac{t}{2}}sin(2t) [/itex]

    which turns into

    [itex]5 \int e^{\frac{t}{2}}sin(2t) = 10sin(2t)e^{\frac{t}{2}}+40cos(2t)e^{\frac{t}{2}} -80\int e^{\frac{t}{2}}sin(2t) [/itex]
    ?
     
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