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Math Help - Integral using Trig Substitution

  1. #1
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    Integral using Trig Substitution

    I'm trying to integrate (1-x^2)^(3/2), however I'm finding it prety difficult. What's the trick? I'm pretty sure you could let x=sint so dx=cost(dt) but I'm not sure if you're allowed to make (1-x^2)^(3/2)=cost and then use the double angle formulas. All help is appreciated!
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    Quote Originally Posted by Dudealadude View Post
    I'm trying to integrate (1-x^2)^(3/2), however I'm finding it prety difficult. What's the trick? I'm pretty sure you could let x=sint so dx=cost(dt) but I'm not sure if you're allowed to make (1-x^2)^(3/2)=cost and then use the double angle formulas. All help is appreciated!
    (\cos^2{t})^{3/2}=\cos^{6/2}{t}=\cos^3{t}
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    Quote Originally Posted by Dudealadude View Post
    I'm trying to integrate (1-x^2)^(3/2), however I'm finding it prety difficult. What's the trick? I'm pretty sure you could let x=sint so dx=cost(dt) but I'm not sure if you're allowed to make (1-x^2)^(3/2)=cost and then use the double angle formulas. All help is appreciated!
    Use the substitution x= cos(t) so that, as dwsmith says, (1- x^2)^{3/2}= (1- cos^2(t))^{3/2}= (sin^2(t))^{3/2}= sin^3(t). And, of course, dx= d(cos(t))= -sin(t)dt.
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