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Math Help - Integration problem

  1. #1
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    Integration problem

    how to integrate (4-x^2)^(-3/2)?
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by hkdrmark View Post
    how to integrate (4-x^2)^(-3/2)?
    \int (4-x^2)^{-3/2} ~dx


    Substitute:
     2\sin(a) = x
     2\cos(a) ~da = dx


    = 2\int cos(a) ~ (4-(2\sin(a))^2)^{-3/2} ~da

    = 2\int cos(a) ~ (4-4\sin^2(a))^{-3/2} ~da

    = 2\int cos(a) ~ 4^{-3/2} ~ (1-\sin^2(a))^{-3/2} ~da

    = 2\int cos(a) ~ \frac 18 ~ (\cos^2(a))^{-3/2} ~da

    = \frac 14 \int cos(a) ~ \sec^3(a) ~da

    = \frac 14 \int \sec^2(a) ~da

    = \frac 14 \tan(a) + C


    anti-substitute:
    a = \arcsin(x/2)


    = \frac 14 \tan(\arcsin(x/2)) + C

    = \frac x{4\sqrt{4-x^2}} + C
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  3. #3
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    thankyou
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  4. #4
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    It's a classic integral, a reciprocal substitution also works.
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  5. #5
    Super Member angel.white's Avatar
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    Quote Originally Posted by Krizalid View Post
    It's a classic integral, a reciprocal substitution also works.
    Could you explain this? I (no joke) thought to myself "What would Krizalid do?" and so I tried making it [(2+x)(2-x)]^(-3/2) and substituting a=2+x, but that got me [a(4+a)]^(-3/2) which didn't seem to get any simpler to me, so I went with this method. I want to see what you're talking about though, I find your methods helpful.
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  6. #6
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    Quote Originally Posted by angel.white View Post
    Could you explain this? I (no joke) thought to myself "What would Krizalid do?" and so I tried making it [(2+x)(2-x)]^(-3/2) and substituting a=2+x, but that got me [a(4+a)]^(-3/2) which didn't seem to get any simpler to me, so I went with this method. I want to see what you're talking about though, I find your methods helpful.
    Let x = \frac{1}{u} so that \frac{dx}{du} = -\frac{1}{u^2} \Rightarrow dx = - \frac{du}{u^2}.

    Then I = - \int (4u^2 - 1)^{-3/2} \, u \, du and the solution easily follows.
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