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Thread: Stuck @ integral :S

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

    hello
    I really stucked at this one

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

    Substitute $\displaystyle x=cos(\theta)$

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

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

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

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

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

    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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    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
    $\displaystyle \int arctan\left( \sqrt{\frac{1-x}{1+x}} \right) dx$
    $\displaystyle \displaystyle\int\frac{dt}{(1+t^2)^2}=\frac12\left (\frac t{t^2+1}+\arctan t\right)+k_1,$ now in your integral substitute $\displaystyle x=\dfrac{1-t^{2}}{1+t^{2}}$ so it becomes $\displaystyle -4\displaystyle\int{\frac{t\arctan t}{\left( t^{2}+1 \right)^{2}}\,dt}=-4\int{\left( -\frac{1}{2\left( 1+t^{2} \right)} \right)'\arctan t\,dt}$ which is equal to $\displaystyle \displaystyle\frac{2\arctan t}{1+t^2}-2\int\frac{dt}{(1+t^2)^2}$ and problem solved.
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