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Math Help - Evaluate this Integral

  1. #1
    Junior Member
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    Evaluate this Integral

    This was a test problem:
    \int^{1}_0\sqrt{\frac{1-x}{x}}dx
    I tried to stare it down until I figured out a solution. I thought about trig substitution, inverse trig, but in the end went with spliting the inside by
    \sqrt{\frac{1}{x}-\frac{x}{x}}}dx
    ran out of time and left it like that....

    hints please?
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  2. #2
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    Hello, ugkwan!

    This is a very messy problem . . . for a test question.


    \displaystyle \int^{1}_0\sqrt{\frac{1-x}{x}}dx

    \displaystyle \text{Let: }\:u \:=\:\sqrt{\frac{1-x}{x}} \quad\Rightarrow\quad u^2 \:=\:\frac{1-x}{x} \quad\Righgtarrow\quad u^2x \:=\:1-x

    . . \displaystyle u^2x + x \:=\:1 \quad\Rightarrow\quad (u^2+1)x \:=\:1 \quad\Rightarrow\quad x \:=\:\frac{1}{u^2+1}<br />

    . . \displaystyle x \:=\:(u^2+1)^{\text{-}1} \quad\Rightarrow\quad dx \:=\:-(u^2+1)^{\text{-}2}\,2u\,du \;=\;\frac{-2u\,du}{(u^2+1)^2}


    \displaystyle\text{Substitute: }\;\int u\cdot\frac{-2u\,du}{(u^2+1)^2} \;=\;-2\int\frac{u^2}{(u^2+1)^2}\,du .[1]


    \displaystyle \text{Partial Fractions: }\;\frac{u^2}{(u^2+1)^2} \;=\;\frac{Ax+B}{u^2+1} + \frac{Cx+D}{(u^2+1)^2}

    . . \text{and we get: }\;A = 0,\;B = 1,\;C = 0,\;D = -1


    \displaystyle \text{And we have: }\;-2\int\left(\frac{1}{u^2+1} - \frac{1}{(u^2+1)^2}\right)\,du

    Can you finish it?


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    I just noticed that [1] can be done "by parts" . . .


    . . \begin{array}{cccccccc}<br />
U &=& u && dV &=& u(u^2+1)^{-2}du \\ \\[-3MM]<br />
dU &=& du && V &=& -\frac{1}{2}{(u^2+1)^{-1}} \end{array}


    \displaystyle \text{And we have: }\;-2\left(-\tfrac{1}{2}\,\frac{u}{u^2+1} + \tfrac{1}{2}\int\frac{du}{u^2+1}\right)
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  3. #3
    Super Member
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    We can use the substitution  x = \sin^2(y)

     dx = 2\sin(y) \cos(y) dy


    The integral becomes :

     \int_0^{\frac{\pi}{2}} \sqrt{\frac{\cos^2(y)}{\sin^2(y)} } ~2\sin(y) \cos(y)~ dy

     \int_0^{\frac{\pi}{2}} ( 1 + \cos(2y) )~dy

     = \left[ y + \frac{\sin(2y) } {2} \right]_0^{\frac{\pi}{2} }

    = \frac{\pi}{2}
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