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Math Help - trig sub integral?

  1. #1
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    trig sub integral?

    so, the problem is integral sq rt (1-4x^2)dx

    i really didn't know how to start, other than the fact that it looks sort of like the a^2-x^2 form of trig sub (in which i think you're supposed to make x= asintheta)

    so, i tried that, but i got lost/confused not long after

    please help!
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  2. #2
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    Hello, buttonbear!

    You have the right idea . . . just carry it through.


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

    Let: . 2x = \sin\theta \quad\Rightarrow\quad x = \tfrac{1}{2}\sin\theta \quad\Rightarrow\quad dx = \tfrac{1}{2}\cos\theta\,d\theta


    Substitute: . \int(\cos\theta)\left(\tfrac{1}{2}\cos\theta\,d\th  eta\right) \;=\;\tfrac{1}{2}\int\cos^2\!\theta\,d\theta \;=\;\tfrac{1}{2}\int\frac{1+\cos2\theta}{2}\,d\th  eta \;=\;\tfrac{1}{4}\int(1 + \cos2\theta)\,d\theta


    And we have: . \tfrac{1}{4}\left(\theta + \tfrac{1}{2}\sin2\theta\right) + C \;=\;\tfrac{1}{4}\left(\theta + \tfrac{1}{2}[2\sin\theta\cos\theta]\right) + C

    . . . . . . . . . = \;\tfrac{1}{4}(\theta + \sin\theta\cos\theta) + C


    Back-substitute: . \sin\theta \:=\:\frac{2x}{1} \:=\:\frac{opp}{hyp}

    . . Then: . adj \,=\,\sqrt{1-4x^2} \quad\Rightarrow\quad \cos\theta \:=\:\sqrt{1-4x^2}


    Theefore: . \frac{1}{4}\left(\arcsin2x + 2x\sqrt{1-4x^2}\right) + C

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  3. #3
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    thanks so much for your help..would you mind just explaining where that first cos comes from? in the integral?
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    Quote Originally Posted by buttonbear View Post
    thanks so much for your help..would you mind just explaining where that first cos comes from? in the integral?
    When substituting x=a\sin\theta, the radical \sqrt{a^2-x^2} becomes a\cos\theta. Here is why:

    \sqrt{a^2-x^2}

    =\sqrt{a^2-(a\sin\theta)^2}

    =\sqrt{a^2-a^2\sin^2\theta}

    =\sqrt{a^2(1-\sin^2\theta)}

    =\sqrt{a^2\cos^2\theta}

    =a\cos\theta (assuming a\geq0 and a suitable restriction on \theta)
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