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Math Help - Stuck @ integral :S

  1. #1
    Member Miss's Avatar
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    Stuck @ integral :S

    hello
    I really stucked at this one

    \int arctan\left( \sqrt{\frac{1-x}{1+x}} \right) dx
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  2. #2
    Super Member General's Avatar
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    Let I=\int arctan\left( \sqrt{\frac{1-x}{1+x}} \right) dx

    Substitute x=cos(\theta)

    \implies \frac{1-x}{1+x}=\frac{1-cos(\theta)}{1+cos(\theta)}

    =\frac{2 \, sin^2(\frac{\theta}{2})}{2 \, cos^2(\frac{\theta}{2})}

    Hence,
    \sqrt{\frac{1-x}{1+x}}=tan\left(\frac{\theta}{2}\right)

    So, I=-\int arctan\left( tan\left(\frac{\theta}{2}\right) \right) \, sin(\theta) \, d\theta

    =\frac{-1}{2}\int \theta \, sin(\theta) \, d\theta

    A quick integration by parts will finish it ..
    Last edited by General; May 14th 2010 at 05:59 AM.
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  3. #3
    Member Miss's Avatar
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    Perfect!!

    thanks
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  4. #4
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    Wait. how did you know substituting cosx for x will solve the problem?
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  5. #5
    Super Member General's Avatar
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    By my mind.
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  6. #6
    Member Miss's Avatar
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    Thanks.
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  7. #7
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    Quote Originally Posted by lilaziz1 View Post
    Wait. how did you know substituting cosx for x will solve the problem?
    He's just very, very good at this!
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  8. #8
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    Quote Originally Posted by Miss View Post
    \int arctan\left( \sqrt{\frac{1-x}{1+x}} \right) dx
    \displaystyle\int\frac{dt}{(1+t^2)^2}=\frac12\left  (\frac t{t^2+1}+\arctan t\right)+k_1, now in your integral substitute x=\dfrac{1-t^{2}}{1+t^{2}} so it becomes -4\displaystyle\int{\frac{t\arctan t}{\left( t^{2}+1 \right)^{2}}\,dt}=-4\int{\left( -\frac{1}{2\left( 1+t^{2} \right)} \right)'\arctan t\,dt} which is equal to \displaystyle\frac{2\arctan t}{1+t^2}-2\int\frac{dt}{(1+t^2)^2} and problem solved.
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