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Thread: Can't evaluate this integral

  1. #1
    Member Em Yeu Anh's Avatar
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    Question Can't evaluate this integral

    $\displaystyle \int_0^{\pi}e^{cost}sin(2t)dt $

    Work I have:
    $\displaystyle Let u = e^{cost}$

    $\displaystyle du=(-e^{cost}sint)dt$

    But I don't know what to do from here since it is a sin(2t) in the original integral.

    Thanks in advance!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Em Yeu Anh View Post
    $\displaystyle \int_0^{\pi}e^{cost}sin(2t)dt $

    Work I have:
    $\displaystyle Let u = e^{cost}$

    $\displaystyle du=(-e^{cost}sint)dt$

    But I don't know what to do from here since it is a sin(2t) in the original integral.

    Thanks in advance!
    sin(2t) = 2sin(t)cos(t)

    do a substitution, u = cos(t), finish off with integration by parts
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by Em Yeu Anh View Post
    $\displaystyle \int_0^{\pi}e^{cost}sin(2t)dt $

    Work I have:
    $\displaystyle Let u = e^{cost}$

    $\displaystyle du=(-e^{cost}sint)dt$

    But I don't know what to do from here since it is a sin(2t) in the original integral.

    Thanks in advance!
    $\displaystyle \sin(2t) =2\sin(t)\cos(t)$ then

    $\displaystyle 2\int \sin(t)\cos(t)e^{\cos(t)}dt$

    Set $\displaystyle u=\cos(t) \implies du =-\sin(t)dt$ to get

    $\displaystyle -2 \int ue^{u}du$ Now use integration by parts

    edit too slow
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  4. #4
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    Quote Originally Posted by Em Yeu Anh View Post
    $\displaystyle \int_0^{\pi}e^{cost}sin(2t)dt $

    Work I have:
    $\displaystyle Let u = e^{cost}$

    $\displaystyle du=(-e^{cost}sint)dt$

    But I don't know what to do from here since it is a sin(2t) in the original integral.

    Thanks in advance!
    $\displaystyle \sin{2t} = 2\sin{t}\cos{t}$.


    So $\displaystyle \int{e^{\cos{t}}\sin{2t}\,dt} = \int{2\sin{t}\cos{t}\,e^{\cos{t}}\,dt}$

    $\displaystyle = -2\int{-\sin{t}\,e^{\cos{t}}\cdot \cos{t}\,dt}$.


    Now use integration by parts...

    $\displaystyle u = \cos{t}$ so $\displaystyle du = -\sin{t}$

    $\displaystyle dv = -\sin{t}\,e^{\cos{t}}$ so $\displaystyle v = e^{\cos{t}}$.


    So $\displaystyle -2\int{-\sin{t}\,e^{\cos{t}}\cdot \cos{t}\,dt} = -2\left[\cos{t}\,e^{\cos{t}} - \int{-\sin{t}\,e^{\cos{t}}\,dt}\right]$

    $\displaystyle = -2\left[\cos{t}\,e^{\cos{t}} - e^{\cos{t}}\right] + C$

    $\displaystyle = 2e^{\cos{t}} - 2\cos{t}\,e^{\cos{t}} + C$.
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  5. #5
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    Quote Originally Posted by Em Yeu Anh View Post
    $\displaystyle \int_0^{\pi}e^{cost}sin(2t)dt $

    Work I have:
    $\displaystyle Let u = e^{cost}$

    $\displaystyle du=(-e^{cost}sint)dt$

    But I don't know what to do from here since it is a sin(2t) in the original integral.

    Thanks in advance!

    $\displaystyle sin(2t) = 2sin(t)cos(t)$

    and $\displaystyle sin(t) = u, cos(t)dt = du$

    so your equation now is:

    $\displaystyle \int_0^{\pi}-e^{u}udu $

    take the negative sign out of the integrand, and solve the integral using integration by parts:

    $\displaystyle u = u, du = du, dv = e^{u}, v = e^{u}$



    can you do it now?
    Last edited by Creebe; Jan 17th 2010 at 06:56 PM.
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  6. #6
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    Quote Originally Posted by Em Yeu Anh View Post
    $\displaystyle \int_0^{\pi}e^{cost}sin(2t)dt $

    Work I have:
    $\displaystyle Let u = e^{cost}$

    $\displaystyle du=(-e^{cost}sint)dt$

    But I don't know what to do from here since it is a sin(2t) in the original integral.

    Thanks in advance!

    $\displaystyle \int\limits_0^\pi e^{\cos t}\sin 2t\,dt=$ $\displaystyle 2\int\limits_0^\pi e^{\cos t}\sin t\cos t\,dt=$ (integration by parts) $\displaystyle 2\left(\left[-\cos t\, e^{\cos t}\right]_0^\pi+\int\limits_0^\pi e^{\cos t}\sin t\,dt\right)$ .

    I think the answer is $\displaystyle 4e$

    Tonio
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  7. #7
    Member Em Yeu Anh's Avatar
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    Wow so many responses! =) Thanks to everyone for helping me out I solved it.
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