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Thread: Integration

  1. #1
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    Integration

    How would integrate the following?



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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by atwinix View Post
    How would integrate the following?


    $\displaystyle \int x^3 (1-x^2)^{-\frac{1}{2}}dx=\int \frac{x^3}{(1-x^2)^{\frac{1}{2}}}dx$

    Let
    $\displaystyle u=(1-x^2)^{\frac{1}{2}}$
    $\displaystyle du =\frac{-x dx}{(1-x^2)^{\frac{1}{2}}}$
    and $\displaystyle x^2 = 1-u^2$

    so
    $\displaystyle \int x^3 (1-x^2)^{-\frac{1}{2}}dx=\int -(1-u^2)du=\frac{u^3}{3} - u = \frac{(1-x^2)^{\frac{3}{2}}}{3} - (1-x^2)^{\frac{1}{2}}$
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  3. #3
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    Write,
    $\displaystyle \frac{1}{2} \int x^2 (1-x^2)^{-1/2} (2x) dx$.
    Let $\displaystyle t=x^2$ to get,
    $\displaystyle \frac{1}{2} \int t(1-t)^{-1/2} dt$
    This is the Beta integral.
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  4. #4
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    It's not the most efficient, but you can use trig sub. I like trig sub.

    A lot of times there's an easier sub to make, but waht the heck.

    $\displaystyle \int\frac{x^{3}}{\sqrt{1-x^{2}}}dx$

    Let $\displaystyle x=sin{\theta}, \;\ dx=cos{\theta}d{\theta}$

    $\displaystyle \int\frac{sin^{3}{\theta}}{\sqrt{1-sin^{2}{\theta}}}cos{\theta}d{\theta}$

    Yoiu can see the identity in the radical: $\displaystyle 1-sin^{2}{\theta}=cos^{2}{\theta}$

    So, it whittles down to:

    $\displaystyle \int{sin^{3}{\theta}}d{\theta}=\frac{-sin^{2}{\theta}cos{\theta}}{3}-\frac{2cos{\theta}}{3}$

    Now, to get it back in terms of x, sub in $\displaystyle {\theta}=sin^{-1}x$

    You get:

    $\displaystyle \frac{-x^{2}\sqrt{1-x^{2}}}{3}-\frac{2}{3}\sqrt{1-x^{2}}=\boxed{\frac{-(x^{2}+2)\sqrt{1-x^{2}}}{3}}$
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