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Math Help - Integral of sqrt(9-x^2)

  1. #1
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    Integral of sqrt(9-x^2)

    \int \sqrt {9-x^2} dx

    I got this in my last exam, and came up with:

    u = \sqrt {9-x^2}
    u' = \frac{1}{2} (9-x^2)^{-1/2}
    v' = dx
    v = x

    So I get

    x \sqrt {9-x^2}  -\frac{1}{2} \int x (9-x^2)^{-\frac{1}{2}} dx

    So then I could

    t = 9-x^2
    -\frac{1}{2} dt = x dx

    to get

    x \sqrt {9-x^2}  +\frac{1}{4} \int t^{-\frac{1}{2}} dt

    But the teacher didn't like it. "Wrong choice", she said; and put me a 1.
    Her solution was

    9 \arcsin \frac{x}{3} + \frac{x}{2} \sqrt {9-x^2} + c

    I've been thinking, but can't come up with anything else... and have no idea how to get to the teacher's solution.

    By the way, I'm a highschool student, haven't started university yet.

    Any help?
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by lainmaster View Post
    \int \sqrt {9-x^2} dx

    I got this in my last exam, and came up with:

    u = \sqrt {9-x^2}
    u' = \frac{1}{2} (9-x^2)^{-1/2}
    v' = dx
    v = x

    So I get

    x \sqrt {9-x^2} -\frac{1}{2} \int x (9-x^2)^{-\frac{1}{2}} dx

    So then I could

    t = 9-x^2
    -\frac{1}{2} dt = x dx

    to get

    x \sqrt {9-x^2} +\frac{1}{4} \int t^{-\frac{1}{2}} dt

    But the teacher didn't like it. "Wrong choice", she said; and put me a 1.
    Her solution was

    9 \arcsin \frac{x}{3} + \frac{x}{2} \sqrt {9-x^2} + c

    I've been thinking, but can't come up with anything else... and have no idea how to get to the teacher's solution.

    By the way, I'm a highschool student, haven't started university yet.

    Any help?
    \int \sqrt {9-x^2} dx

    Use the substitution x=3\sin(\theta) \implies dx=3\cos(\theta)d\theta

    \int \sqrt{9-9\sin^2(\theta)}(3\cos(\theta))d\theta

    9\int\\cos(\theta) \sqrt{1-sin^{2}(\theta)}d\theta

    9\int \cos^{2}(\theta)d\theta=\frac{9}{2}\int 1+\cos(2\theta)d\theta=\frac{9}{2}\left( \theta + \frac{1}{2}\sin(2\theta)\right)

    \frac{9}{2}\left( \theta +\sin(\theta)\cos(\theta)\right)=\frac{9}{2}\left( \sin^{-1}\left( \frac{x}{3}\right)+\frac{x}{3}\frac{\sqrt{9-x^2}}{3}\right)=

    \frac{9}{2}\sin^{-1}\left( \frac{x}{3}\right)+\frac{1}{2}x\sqrt{9-x^2}+c
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  3. #3
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    Thanks a lot =)
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  4. #4
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    If it helps at all, the u' you got when doing integration by parts was incorrect. That'd be why you weren't able to pull off something at least similar.
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  5. #5
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    Uh... yes, it helps.
    To be honest, I never learned how to do derivatives.
    Thanks
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  6. #6
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    Hey, I never learned u-substitution. =p But we get by.
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