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    Hopefully somebody will be able to help me with this.
    For some reason I am ending up with 0=0 for an answer, I think it may be me getting signs mixed up somewhere.
    Here is my question and workings out:-
     \int e^xSin2xdx
    I=\int e^xSin2xdx
    let u=e^x
    \frac {du}{dx}=e^x
    \frac {dv}{dx}Sin2x
    v=- \frac{1}{2}Cos2x
    I=e^x*-\frac {1}{2}Cos2x - \int-\frac{1}{2}Cos2x*e^xdx
    I=-\frac{1}{2}e^xCos2x+\int \frac{1}{2}e^xCos2xdx
    let u=Cos2x
    \frac{du}{dx}=-2Sin2x
    \frac{dv}{dx}=\frac{1}{2}e^x
    v=\frac{1}{2}e^x
    I=-\frac{1}{2}e^xCos2x+\frac{1}{2}e  ^xCos2x-\int\frac{1}{2}e^x-2Sin2xdx
    I=-\frac{1}{2}e^xCos2x+\frac{1}{2}e  ^xCos2x-\int-e^xSin2xdx
    I=-\frac{1}{2}e^xCos2x+\frac{1}{2}e  ^xCos2x+\int e^xSin2xdx
    I=-\frac{1}{2}e^xCos2x+\frac{1}{2}e  ^xCos2x+I
    I+I=-\frac{1}{2}e^xCos2x+\frac{1}{2}e  ^xCos2x
    0=0

    Where am I going wrong?

    Barney
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    you have to stick to using the same functions for u and dv/dx, otherwise...

    (i.e. for the 2nd integration, stick with u=exp term, dv/dx = trig term)
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    PS Helper
    For the second by parts let  u = \frac{1}{2}e^x again and \frac{dv}{dx} = cos2x

    Keep the u and dv/dx the same parts.
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    So using:-
    u=\frac{1}{2}e^x
    \frac{du}{dx}=\frac{1}{2}e^x
    \frac{dv}{dx}=Cos2x
    v=SinxCosx
    I get:-
    I=-\frac{1}{2}e^xCos2x+\frac{1}{2}e  ^xSinxCosx-\int \frac{1}{2}e^xSinxCosxdx
    Where do I get the \int e^xSin2xdx for the substitution of I?

    Barney
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    if \diasplaystyle \frac{dv}{dx}=cos(2x) the,

     \int cos(2x)dx=\frac{1}{2}sin(2x)

    giving the last part as \frac{-1}{4}\int e^{x}sin(2x)

    then - because you have a multiple of the same integral on the LHS (muliply both sides by 4 and you`ll see this), you can put the same one all on one side of the equals sign and...
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    I thought you only did the derivative of u, and that you integrated \frac{dv}{dx}.

    Barney
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    But, i was pointing out that, when you integrated your dv/dx you got Sin(x)Cos(x), when in fact the answer`s (1/2)Sin(2x).

    Hang on.. i`ll type out what you should get (but not the answer, `cause that`s the easy part - you`ve basically got it, but i think the arithmetic is maybe freaking you out!)
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    \int e^{2}sin(2x)dx=\frac{-1}{2}cos(2x)- \frac{-1}{2} \int e^{x}cos(2x)dx

    taking the second part:

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

    giving, together:

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

    multiply this by 4, and take the last term over to the LHS...
    • Thread Starter
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    Sorry for the delay and also potentially being thick, but where did the \int e^2Sin(2x)dx come from?

    Barney
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    !!!!Bump!!!!
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    TSR Support Team
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    (Original post by Barney63)
    Sorry for the delay and also potentially being thick, but where did the \int e^2Sin(2x)dx come from?

    Barney
    From e^22sinxcosx
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    maybe this will explain it better...

    http://www.youtube.com/watch?v=1bSTXcFeaZ4
 
 
 
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