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Math Help - Integration by parts (3x-∏)cos(x/2)

  1. #1
    Junior Member froodles01's Avatar
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    Integration by parts (3x-∏)cos(x/2)

    I know this can be done long hand & several ways.
    I started from integrating by parts for definite integral (as I will be adding the limits later)
    ∫ f(x)g'(x) dx = f(x)g(x) - ∫f'(x)g(x) dx

    Let f(x) = (3x-∏) so f'(x) = 3
    then g'(x) = cos (x/2) & from standard integral table g(x) = -1/2 sin(x/2)

    then I used substitution for ∫-1/2 sin(x/2) dx
    let u = x/2
    so -1/2∫ sin u du = -3/2 cos u +c = -3/2 cos(x/2) +c

    Now ∫ f(x)g'(x) dx = (3x-∏) (-1/2 sin(x/2)) - 3(-3/2 cos(x/2)) + c

    simplifying
    = (3x-∏) (-1/2 sin(x/2))- (1/2(2 cos(x/2)) +c

    Please advise:
    where did i start going wrong?
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  2. #2
    Senior Member
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    Re: Integration by parts (3x-∏)cos(x/2)

    You are wrong with g(x). If g'(x)= cos(x/2) then g(x)=2sin(x/2)
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