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Math Help - Help with integration problem

  1. #1
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    Help with integration problem

    This problem has been worked out for me, but I am confused about what exactly is going on in it.

    Problem: Find the indefinite integral.
    ∫-1/sqrt(1-(2t-1)^2) dt

    Step 1: Let u= 2t -1

    Step 2: du= 2 dt

    Step 3: ∫ -1/sqrt(1-(2t-1)^2) dt
    = -1/2 ∫ 2/sqrt(1-(2t-1)^2) dt

    Step 4: = -1/2 arcsin(2t-1) + C

    The main thing I don't understand is where the
    -1/2 before the ∫ sign is coming from in Step 3, and why the -1 changed to a 2 in the part after the ∫ sign. If anyone could explain to me what is going on in this problem I would greatly appreciate it. Thanks.
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  2. #2
    Newbie Firstone's Avatar
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    Lets say the denominator = x to save me from typing alot.

    ∫ -1/x
    ∫ -1/2 * 2/x

    since -1/2 is a constant you may bring it to the front.
    -1/2 ∫ 2/x


    See how they are the same?
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  3. #3
    Super Member Showcase_22's Avatar
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    Since dt/dx=(1/2)cosx, the 1/2 can be moved to the front of the integral as well as the -1. At the front of the integral this gives -1(1/2) which is -1/2.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Showcase_22 View Post


    Since dt/dx=(1/2)cosx, the 1/2 can be moved to the front of the integral as well as the -1. At the front of the integral this gives -1(1/2) which is -1/2.
    Or you could let u=2t-1\implies \,du=2\,dt

    The integral transforms into -\tfrac{1}{2}\int\frac{\,du}{\sqrt{1-u^2}}, which we know to be -\tfrac{1}{2}\sin^{-1}u+C

    So the final result would be -\tfrac{1}{2}\sin^{-1}(2t-1)+C.

    --Chris
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