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Thread: Trig substitution

  1. #1
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    Trig substitution

    $\displaystyle \int \frac{x}{\sqrt{3-x^4}}dx$

    $\displaystyle \int \frac{x}{\sqrt{\sqrt{3}^2-(x^2)^2}}dx$

    $\displaystyle x = \sqrt{3}sinx$

    $\displaystyle dx = \sqrt{3}cosx$

    $\displaystyle \sqrt{3-(3sin^2\theta)^2}$

    $\displaystyle \sqrt{3-(1 - 3sin^4\theta)}$

    This is where i get stuck.

    Any help is appreciated
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  2. #2
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    Quote Originally Posted by silencecloak View Post
    $\displaystyle \int \frac{x}{\sqrt{3-x^4}}dx$

    [snip]
    If you make the substitution $\displaystyle u = x^2$ you get a standard form ....
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  3. #3
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    doh

    i need to develop an eye for these sorts of things still

    thanks
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