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Math Help - [SOLVED] Integration by substitution

  1. #1
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    [SOLVED] Integration by substitution

    Use the substitution to evaluate

    The problem before this was using the same substitution, but to evaluate instead. I was able to do that one easily, but I don't see how I can use this substitution for this problem.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by ban26ana View Post
    Use the substitution to evaluate

    The problem before this was using the same substitution, but to evaluate instead. I was able to do that one easily, but I don't see how I can use this substitution for this problem.

    Observe that u={\color{blue}1-x^2}\implies {\color{red}x^2}=1-u, then \,du={\color{green}-2x\,dx}. The limits of integration change as well: u(0)=1 and u(1)=0

    So \int_0^1 x^3\sqrt{1-x^2}\,dx=-\tfrac{1}{2}\int_0^1{\color{green}-2x}\cdot {\color{red}x^2}\sqrt{{\color{blue}1-x^2}}{\color{green}\,dx} \xrightarrow{u=1-x^2}{}-\tfrac{1}{2}\int_1^0\left(1-u\right)\sqrt{u}\,du=\tfrac{1}{2}\int_0^1u^{1/2}-u^{3/2}\,du

    Can you continue from here?
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  3. #3
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    Yes I can. Thanks so much!
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  4. #4
    Super Member General's Avatar
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    Quote Originally Posted by ban26ana View Post
    Use the substitution to evaluate

    The problem before this was using the same substitution, but to evaluate instead. I was able to do that one easily, but I don't see how I can use this substitution for this problem.
    Good question.
    You have the substitution u=1-x^2
    so \sqrt{1-x^2}=\sqrt{u}
    The problem here is: you want to find x^3 and dx in terms of u.
    Its easy; solve the substitution for x then differentiate both side with respect to x to get the dx.
    Also find x^3 from your substitution.
    Can you do that?
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