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Math Help - integration problem

  1. #1
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    integration problem

    in (sin(x) cos(x)) dx
    let u=cos(x)
    du=-sin(x) dx
    answer is -1/2cos^2(x) (a)

    if let u=sin(x)
    du=cos(x) dx
    then answer is 1/2*sin^2(x) (b)

    if we put it into an right angle triangle length 3,4,5

    then (a) get -0.18
    but (b) get 0.32
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  2. #2
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    Re: integration problem

    Quote Originally Posted by oponion View Post
    in (sin(x) cos(x)) dx
    let u=cos(x)
    du=-sin(x) dx
    answer is -1/2cos^2(x) (a)

    if let u=sin(x)
    du=cos(x) dx
    then answer is 1/2*sin^2(x) (b)

    if we put it into an right angle triangle length 3,4,5

    then (a) get -0.18
    but (b) get 0.32
    You have forgotten about the arbitrary constant that is added to each integral. Both answers are correct up to a different constant, since \displaystyle \begin{align*} -\frac{1}{2}\cos^2{x} = -\frac{1}{2}\left(1 - \sin^2{x}\right) = -\frac{1}{2} + \frac{1}{2}\sin^2{x} \end{align*}.
    Thanks from oponion
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  3. #3
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    Re: integration problem

    thank you very much
    because my answer is different from the solution
    and i use x=1,2 sub into my answer and the solution and get different answer
    now i understand
    thank you
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