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Thread: hi i need to integrate this.

  1. #1
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    hi i need to integrate this.

    Hi i'm in a bit of a hurry, any help would be great please see the attachment.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by mbbx5va2 View Post
    Hi i'm in a bit of a hurry, any help would be great please see the attachment.
    Question is: $\displaystyle \int^1_0 \frac{1}{\sqrt{x(1-x)}} dx$

    Use$\displaystyle x = sin^2(\theta)$ to get $\displaystyle \int^{\frac{\pi}{2}}_0 1 d\theta$


    $\displaystyle x = sin^2(\theta)$ then our denominator becomes: $\displaystyle \sqrt{sin^2(\theta)(1-sin^2(\theta))} = \sqrt{sin^2(\theta)cos^2(\theta)}$ $\displaystyle = sin(\theta)cos(\theta) $

    We can also use our substitution to change the limits:
    $\displaystyle 1 = sin^2(\theta) \: , \: \theta= \frac{pi}{2}$

    $\displaystyle 0 = sin^2(\theta) \: , \: \theta = 0$

    $\displaystyle \int^{\frac{\pi}{2}}_0 \frac{1}{sin(\theta)cos(\theta)}d\theta$

    Not sure where the 1 came from but that integral can be solved using integration by parts
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  3. #3
    Member Nacho's Avatar
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    Maybe you should work a bit: $\displaystyle x-x^2$

    That seem an arcsin, then:

    $\displaystyle \dfrac{1}{4} + \left( -\dfrac{1}{4}+x-x^2 \right)=\dfrac{1}{4} - \left( x^2-x+\dfrac{1}{4}\right)$ $\displaystyle =\left(\dfrac{1}{2} \right)^2-\left(x-\dfrac{1}{2} \right)^2$ $\displaystyle =
    \left( {\frac{1}
    {2}} \right)^2 \left\{ {1 - \left( {\frac{{x - \frac{1}
    {2}}}
    {{\frac{1}
    {2}}}} \right)^2 } \right\}
    $

    can you see the substitute?
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  4. #4
    Math Engineering Student
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    put $\displaystyle x=\frac{1}{1+t^{2}}$ and your integral becomes $\displaystyle \int_{0}^{\infty }{\frac{2}{1+t^{2}}\,dt}=\pi .$
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  5. #5
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    Question is: $\displaystyle \int^1_0 \frac{1}{\sqrt{x(1-x)}} dx$

    Use$\displaystyle x = sin^2(\theta)$ to get $\displaystyle \int^{\frac{\pi}{2}}_0 1 d\theta$


    $\displaystyle x = sin^2(\theta)$ then our denominator becomes: $\displaystyle \sqrt{sin^2(\theta)(1-sin^2(\theta))} = \sqrt{sin^2(\theta)cos^2(\theta)}$ $\displaystyle = sin(\theta)cos(\theta) $

    We can also use our substitution to change the limits:
    $\displaystyle 1 = sin^2(\theta) \: , \: \theta= \frac{pi}{2}$

    $\displaystyle 0 = sin^2(\theta) \: , \: \theta = 0$

    $\displaystyle \int^{\frac{\pi}{2}}_0 \frac{1}{sin(\theta)cos(\theta)}d\theta$

    Not sure where the 1 came from but that integral can be solved using integration by parts
    This looks fishy... $\displaystyle dx\ne d\theta$.

    If you use $\displaystyle x = \sin^2\theta$, then $\displaystyle dx = 2\sin\theta\cos\theta d\theta$ which will give you

    $\displaystyle \int^{\frac{\pi}{2}}_0 2 d\theta=\pi$ and that's agrees with another imposter's post.
    Last edited by matheagle; May 19th 2009 at 11:07 PM.
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  6. #6
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by matheagle View Post
    This looks fishy... $\displaystyle dx\ne d\theta$.

    If you use $\displaystyle x = \sin^2\theta$, then $\displaystyle dx = 2\sin\theta\cos\theta d\theta$ which will give you

    $\displaystyle \int^{\frac{\pi}{2}}_0 2 d\theta=\pi$ and that's agrees with another imposter's post.
    -facepalm-

    how did I not see that? Thanks for pointing that out
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