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Math Help - [SOLVED] integration using trigonometry substitution

  1. #1
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    Unhappy [SOLVED] integration using trigonometry substitution

    \int_{\frac{1}{2}}^{1}\frac{\sqrt{1-x^2}}{x^2} dx [ using the substitution, x = costheta]

     \frac{dx}{d\theta }= -sin\theta

     dx = -sin\theta d\theta

     \int_{\frac{1}{2}}^{1}\frac{\sqrt{1-cos^2\theta }}{cos^2\theta } dx

    \int_{\frac{1}{2}}^{1}\frac{\sqrt{1-cos^2\theta }}{cos^2\theta } .-sin\theta  dtheta

    \int_{\frac{1}{2}}^{1}\frac{sin\theta }{cos^2\theta }.-sin\theta dtheta

     \int_{\frac{1}{2}}^{1}\frac{- sin^2\theta}{cos^2\theta }

    never mind i got it
    Last edited by differentiate; September 25th 2009 at 08:21 AM. Reason: got the answer
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  2. #2
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    Hello differentiate
    Quote Originally Posted by differentiate View Post
    \int_{\frac{1}{2}}^{1}\frac{\sqrt{1-x^2}}{x^2} dx [ using the substitution, x = costheta]

     \frac{dx}{d\theta }= -sin\theta

     dx = -sin\theta d\theta

     \int_{\frac{1}{2}}^{1}\frac{\sqrt{1-cos^2\theta }}{cos^2\theta } dx

    \int_{\frac{1}{2}}^{1}\frac{\sqrt{1-cos^2\theta }}{cos^2\theta } .-sin\theta  dtheta

    \int_{\frac{1}{2}}^{1}\frac{sin\theta }{cos^2\theta }.-sin\theta dtheta

     \int_{\frac{1}{2}}^{1}\frac{- sin^2\theta}{cos^2\theta }

    x = cos@
    when x = 1,
    @ = 0
    when x = 1/2
    @ = pi/3

    what do I do next?
    The integral becomes \int_{0}^{\frac{\pi}{3}}\tan^2\theta d\theta, which you can solve using \tan^2\theta = \sec^2\theta-1, and noting that \int\sec^2\theta d\theta = \tan\theta.

    Grandad
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