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Math Help - Trig Substitution

  1. #1
    Junior Member
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    Trig Substitution

    Where is my error in this problem?

    \int\sqrt{1-4x^2}\,dx

    x=\frac{1}{2}\sin{t}
    dx=\frac{1}{2}\cos{5}

    \sqrt{1-4x^2}=\sqrt{1-\sin^2{t}}=\sqrt{\cos^2{t}}=|cos t|=cos t

    \int\sqrt{1-4x^2}\,dx
    =\frac{1}{2}\int\cos^2{t}\,dt
    =\frac{1}{4}\int{1+\cos{2t}}\,dt
    =\frac{1}{4}t + \frac{1}{8}\sin{2t}

    Set up triangle with angle t, adjacent =\sqrt{1-4x^2}, opposite=2x, and hypotenuse=1.

    \frac{1}{4}sin^{-1}{2x}+\frac{x\sqrt{1-4x^2}}{8}+C
    Last edited by Mike9182; September 2nd 2010 at 09:02 AM. Reason: latex error
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  2. #2
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    sin(2t)= 2 sin(t) cos(t). Here, you have sin(t)= 2x and, from sin^2(t)+ cos^2(t)= 1, cos(t)= \sqrt{1- sin^2(t)}= \sqrt{1- 4x^2} so that sin(2t)= 2(2x)(\sqrt{1- 4x^2}).

    \frac{1}{8} sin(2t)= \frac{1}{8}\left(4x\sqrt{1- 4x^2}\right)= \frac{1}{2}x\sqrt{1- 4x^2}.
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  3. #3
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    Hello, Mike9182!

    Your error is a silly one in the last step . . . back-substituting.


    \frac{1}{4}t + \frac{1}{8}\sin{2t} + C


    \sin t \:=\:2x

    Set up triangle with angle t\!:\;\;adj =\sqrt{1-4x^2},\;opp=2x,\;hyp=1

    . . This is all correct!

    So we have: . \begin{Bmatrix}\sin t \:=\:2x \\ \cos t \:=\:\sqrt{1-4x^2} \end{Bmatrix}

    Then: . \sin 2t \:=\:2\sin t\cos t \:=\:2(2x)(\sqrt{1-4x^2}) \;=\;4x\sqrt{1-4x^2}


    Substitute: . \frac{1}{4}\sin^{-1}(2x) + \frac{1}{8}\left(4x\sqrt{1-4x^2}\right) + C

    . . . . . . . =\;\frac{1}{4}\sin^{-1}(2x) + \frac{1}{2}x\sqrt{1-4x^2} + C

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