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Math Help - Integration by parts query. Getting 0 as answer, think its wrong, help!!

  1. #1
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    Integration by parts query. Getting 0 as answer, think its wrong, help!!

    Hi,

    I'm asked to integrate e^2xcosxdx. I have used intefration by parts letting u=e^2x and dv=cosxdx. This results in

    e^2xsinx-the integral of 2sinxe^2x. I then integrated by parts again and the overall result was if I = integral of e^2xcosxdx then

    I= e^2xsinxdx-e^2xsinxdx + 0.5( I)

    0.5I = 0

    Therefore I=0
    Is this wrong?
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  2. #2
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    Re: Integration by parts query. Getting 0 as answer, think its wrong, help!!

    Quote Originally Posted by orlacoon View Post
    Hi,

    I'm asked to integrate e^2xcosxdx. I have used intefration by parts letting u=e^2x and dv=cosxdx. This results in

    e^2xsinx-the integral of 2sinxe^2x. I then integrated by parts again and the overall result was if -I = integral of e^2xcosxdx then

    I= e^2xsinxdx-e^2xsinxdx + 0.5( I)

    0.5I = 0Therefore I=0
    Is this wrong?
    You are making a sign error.
    Add that last integral to both sides and then divide by 2.
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  3. #3
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    Re: Integration by parts query. Getting 0 as answer, think its wrong, help!!

    Still not sure where I'm going wrong. Could you explain further? Thanks
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  4. #4
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    Re: Integration by parts query. Getting 0 as answer, think its wrong, help!!

    You should end up with the same integration but with a minus sign on the right side. You move that integration to the left hand side and then divide by two.

    So you should end up with something like:

    I= .......... -I (move right hand integration to left hand side)

    2I = .............. (divide by two)
    I = 1/2(..........) (Final answer)

    Sorry i didn't do the integration. In class right now =p
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    Re: Integration by parts query. Getting 0 as answer, think its wrong, help!!

    Quote Originally Posted by Plato View Post
    You are making a sign error.
    Add that last integral to both sides and then divide by 2.
    He's right. Remember to pick the right u and v' elements.
    Derivative of (sinx)' = cosx and (cosx)' = -sinx
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  6. #6
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    Re: Integration by parts query. Getting 0 as answer, think its wrong, help!!

    \int e^{2x}\cos{x} \, dx

    u = \cos{x} , du = -\sin{x} \, dx

    dv = e^{2x} \, dx , v = \frac{1}{2}e^{2x}

    \int e^{2x}\cos{x} \, dx = \frac{1}{2}e^{2x}\cos{x} + \frac{1}{2}\int e^{2x} \sin{x} \, dx

    p = \sin{x} , dp = \cos{x} \, dx

    dq = e^{2x} \, dx , q = \frac{1}{2}e^{2x}

    \int e^{2x}\cos{x} \, dx = \frac{1}{2}e^{2x}\cos{x} + \frac{1}{2}\left[\frac{1}{2}e^{2x}\sin{x} - \frac{1}{2}\int e^{2x}\cos{x} \, dx\right]

    \int e^{2x}\cos{x} \, dx = \frac{1}{2}e^{2x}\cos{x} + \frac{1}{4}e^{2x}\sin{x} - \frac{1}{4}\int e^{2x}\cos{x} \, dx

    \frac{5}{4} \int e^{2x}\cos{x} \, dx = \frac{1}{2}e^{2x}\cos{x} + \frac{1}{4}e^{2x}\sin{x} + C

    \int e^{2x}\cos{x} \, dx = \frac{2}{5}e^{2x}\cos{x} + \frac{1}{5}e^{2x}\sin{x} + C
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  7. #7
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    Re: Integration by parts query. Getting 0 as answer, think its wrong, help!!

    Thanks a lot, got it wrong way around!
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