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    ∫2e^(x).Sin(x)Cos(x) dx

    v= Sin(x)Cos(x)
    dv/dx = Cos^2 (x) - Sin^2 (x) = Cos2(x)

    du/dx = 2e^(x)
    u= 2e^(x)

    ∫2e^(x).Sin(x)Cos(x) dx = 2e^(x) . Sin(x)Cos(x) - ∫ 2e^(x). Cos2(x)

    v=Cos2(x)
    dv/dx = -2sin2(x)

    du/dx= 2e^(x)
    u= 2e^(x)

    ∫2e^(x).Sin(x)Cos(x) dx = 2e^(x).Sin(x)Cos(x) - [ 2e^(x).Cos2(x) - ∫2e^(x).-2sin2(x) dx]
    =>
    ∫2e^(x).Sin(x)Cos(x) dx = 2e^(x).Sin(x)Cos(x) - [2e^(x).Cos2(x) + 2∫2e^(x).sin2(x) dx]
    =>
    ∫2e^(x).Sin(x)Cos(x) dx = 2e^(x).Sin(x)Cos(x) - 2e^(x).Cos2(x) - 2∫2e^(x).sin2(x) dx

    rearranging this whole mess
    ∫2e^(x).Sin(x)Cos(x) + 2∫2e^(x).sin2(x) dx = 2e^(x).Sin(x)Cos(x) - 2e^(x).Cos2(x)

    3∫2e^(x).Sin(x)Cos(x) dx = e^(x).(sin2(x)-2Cos2(x)
    textbook answer is (e^(x) /5).(sin2(x)-2Cos2(x)
    where does the 5 come from? can anyone spot the mistake please
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    do you still need an answer?
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    yes please
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    have you got like a Snapchat or messenger because tsr isn't letting me upload a photo of the working out
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    (Original post by MikeyA)
    have you got like a Snapchat or messenger because tsr isn't letting me upload a photo of the working out
    It would be better if you can find the OPs mistake if possible instead of posting your own working.
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    okee
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    To Op, I think it would be easier for you to simplify the integral in question, preferably using sin2x = 2sinxcosx so you can remove sinxcosx from your integral.

    It would also help to keep your numbers outside the integral.
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    (Original post by Carlos Nim)
    x
    Your mistake is going from here

    ∫2e^(x).Sin(x)Cos(x) + 2∫2e^(x).sin2(x) dx

    to here

    3∫2e^(x).Sin(x)Cos(x) dx

    Please try again and if you get the same thing post your working in full for this step.

    As the poster above said, it would have been much easier if you changed sin(x)cos(x) using the identity before doing any integration and you probably wouldn't have made this mistake.
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    ∫2e^(x).Sin(x)Cos(x) dx = 2e^(x).Sin(x)Cos(x) - [ 2e^(x).Cos2(x) - ∫2e^(x).-2sin2(x) dx]
    =>
    ∫2e^(x).Sin(x)Cos(x) dx = 2e^(x).Sin(x)Cos(x) - [2e^(x).Cos2(x) + ∫4e^(x).sin2(x) dx]
    =>
    ∫2e^(x).Sin(x)Cos(x) dx = 2e^(x).Sin(x)Cos(x) - 2e^(x).Cos2(x) - 4∫e^(x).sin2(x) dx]
    =>
    ∫2e^(x).Sin(x)Cos(x) + 4∫e^(x).sin2(x) dx = 2e^(x).Sin(x)Cos(x) - 2e^(x).Cos2(x)
    =>
    6∫e^(x).Sin(x)Cos(x) dx = e^(x).(sin2(x)-2Cos2(x)
    =>
    ∫e^(x).Sin(x)Cos(x) dx = e^(x)/6.(sin2(x)-2Cos2(x)

    Im not sure about this, anyway i got 6 , the answer required is 5
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    I would suggest to start again and simplify the integral first using the identity I gave earlier, it would make the simplification at the later stages much easier and this you would be less likely to mess up
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