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Math Help - trig. substitution

  1. #1
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    trig. substitution

    \int x \sqrt{16-4x^2}dx    =>   \int x \sqrt{4^2-(2x)^2}dx

    2x = 4sin\theta

    2dx = 4cos\theta d\theta

    dx = 2cos\theta d\theta

    = \int x \sqrt{4^2-(4sin\theta)^2}*2cos\theta d\theta

    = 2\int x \sqrt{16(1-sin^2\theta)}*cos\theta d\theta

    = 2 \int x(4cos\theta cos\theta)d\theta

    = 2 \int x(4cos^2\theta)d\theta

    According to my substitution, should I let x=2sin \theta and substitute that back in here??

    = 2 \int 2sin\theta(4cos^2d\theta) ???

    If so, where should I go from there?
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  2. #2
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    <br /> <br />
= 16 \int sin\theta(cos^2\theta d\theta)<br />

    use u substitution and let u=cos \theta

    after integration use your initial supposition that  2x = 4sin\theta to rewrite your answer in terms of x.
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  3. #3
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    2 \int 2sin\theta(4cos^2\theta)d\theta = 16 \int sin\theta(cos^2\theta)d\theta

    = 16 \left[ {\frac{cos^3\theta}{3}} \right] +c

    (Check it works by differentiating back.) In general:

    \int f'(x)f^n(x) dx = \frac{f^{n+1}(x)}{n+1} +c
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  4. #4
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    how do I rewrite:
    \frac{-16cos^3\theta}{3} +C
    in terms of x? I don't understand.
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  5. #5
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    Hello, saiyanmx89!

    How do I rewrite: -\frac{16}{3}\cos^3\!\theta +C in terms of x ?
    Is this your first Trig Substitution?


    We had: . 2x  \:=\:4\sin\theta \quad\Rightarrow\quad \sin\theta \:=\:\frac{x}{2} \:=\:\frac{opp}{hyp}

    Using Pythagorus, we find that: . adj \:=\:\sqrt{4-x^2}

    . . Hence: . \cos\theta \:=\:\frac{\sqrt{4-x^2}}{2}


    Substitute: . -\frac{16}{3}\cos^3\!\theta + C \;=\;-\frac{16}{3}\left(\frac{\sqrt{4-x^2}}{2}\right)^3 + C \;=\;-\frac{16}{3}\cdot\frac{\left(\sqrt{4-x^2}\right)^3}{8} + C

    . . . . . . . = \;-\frac{2}{3}(4-x^2)^{\frac{3}{2}} + C

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