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Math Help - 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.
    Attached Files Attached Files
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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: \int^1_0 \frac{1}{\sqrt{x(1-x)}} dx

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


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

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

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

    \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: x-x^2

    That seem an arcsin, then:

    \dfrac{1}{4} + \left( -\dfrac{1}{4}+x-x^2 \right)=\dfrac{1}{4} - \left( x^2-x+\dfrac{1}{4}\right) =\left(\dfrac{1}{2} \right)^2-\left(x-\dfrac{1}{2} \right)^2 =<br />
\left( {\frac{1}<br />
{2}} \right)^2 \left\{ {1 - \left( {\frac{{x - \frac{1}<br />
{2}}}<br />
{{\frac{1}<br />
{2}}}} \right)^2 } \right\}<br />

    can you see the substitute?
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    put x=\frac{1}{1+t^{2}} and your integral becomes \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: \int^1_0 \frac{1}{\sqrt{x(1-x)}} dx

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


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

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

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

    \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... dx\ne d\theta.

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

     \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... dx\ne d\theta.

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

     \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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