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Math Help - integration problem

  1. #1
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    integration problem

    prove that ... integrate sqrt [(1-x)/(1+x)] dx ( upper limit = 1 , lower limit =0 ) = pi/2 - 1

    by using x = cos 2 theta

    thanks in advanced!
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  2. #2
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    Quote Originally Posted by ose90 View Post
    prove that ... integrate sqrt [(1-x)/(1+x)] dx ( upper limit = 1 , lower limit =0 ) = pi/2 - 1

    by using x = cos 2 theta

    thanks in advanced!
    Note:

    1. \frac{1 - \cos (2 \theta)}{1 + \cos (2 \theta)} = \frac{1 - \cos^2 \theta + \sin^2 \theta}{1 + \cos^2 \theta - \sin^2 \theta} = \frac{2 \sin^2 \theta}{2 \cos^2 \theta} = \tan^2 \theta.

    2. dx = -2 \sin (2 \theta) \, d \theta = -4 \sin \theta \cos \theta \, d \theta.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Note:

    1. \frac{1 - \cos (2 \theta)}{1 + \cos (2 \theta)} = \frac{1 - \cos^2 \theta + \sin^2 \theta}{1 + \cos^2 \theta - \sin^2 \theta} = \frac{2 \sin^2 \theta}{2 \cos^2 \theta} = \tan^2 \theta.

    2. dx = -2 \sin (2 \theta) \, d \theta = -4 \sin \theta \cos \theta \, d \theta.

    Thanks you very much, I wrongly conceive that 1-cos 2 theta = sin theta and 1+ cos 2 theta = cos theta , no wonder I couldn't solve it!

    Now I can solve it, thank you!
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    Note:

    1. \frac{1 - \cos (2 \theta)}{1 + \cos (2 \theta)} = \frac{1 - \cos^2 \theta + \sin^2 \theta}{1 + \cos^2 \theta - \sin^2 \theta} = \frac{2 \sin^2 \theta}{2 \cos^2 \theta} = \tan^2 \theta.

    2. dx = -2 \sin (2 \theta) \, d \theta = -4 \sin \theta \cos \theta \, d \theta.

    Hmm, I have solved it ! Thanks if you have come here again.
    Last edited by ose90; November 17th 2008 at 04:52 AM.
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