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Integration Problem

  1. May 12, 2005 #1
    Show that

    [tex]\int _{\sqrt{2}} ^2 \frac{1}{t^3 \sqrt{t^2 - 1}} \: dt = \frac{1}{24}\left( -6 + 3\sqrt{3} + \pi \right)[/tex]

    I've tried to use the trigonometric substitution [tex]t=\sec \theta[/tex], but without success thus far.

    [tex]t=\sec \theta \Rightarrow \frac{dt}{d\theta}=\sec \theta \tan \theta \Rightarrow dt = \sec \theta \tan \theta \: d\theta[/tex]

    [tex]\int _{\sqrt{2}} ^{2} \frac{1}{t^3 \sqrt{t^2 - 1}} \: dt=\int _{\sec \sqrt{2}} ^{\sec 2} \frac{\sec \theta \tan \theta}{\sec ^3 \theta \sqrt{\sec ^2 \theta - 1}} \: d\theta[/tex]

    [tex]\int _{\sec \sqrt{2}} ^{\sec 2} \frac{\sec \theta \tan \theta}{\sec ^3 \theta \sqrt{\sec ^2 \theta - 1}} \: d\theta = \int _{\sec \sqrt{2}} ^{\sec 2} \frac{\tan \theta}{\sec ^2 \theta \sqrt{\tan ^2 \theta}} \: d\theta[/tex]

    [tex]\int _{\sec \sqrt{2}} ^{\sec 2} \frac{\tan \theta}{\sec ^2 \theta \sqrt{\tan ^2}} \: d\theta = \int _{\sec \sqrt{2}} ^{\sec 2} \frac{d\theta}{\sec ^2 \theta} = \int _{\sec \sqrt{2}} ^{\sec 2} \cos ^2 \theta \: d\theta[/tex]

    [tex]\int _{\sec \sqrt{2}} ^{\sec 2} \cos ^2 \theta \: d\theta = \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} (1+\cos 2\theta) \: d\theta = \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} \: d\theta + \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} \cos 2\theta \: d\theta[/tex]

    Consider the following

    [tex]\int \cos 2x \: dx[/tex]

    [tex]u=2x \Rightarrow \frac{du}{dx}= 2 \Rightarrow dx = \frac{du}{2}[/tex]

    [tex]\int \cos 2x \: dx = \frac{1}{2} \int \cos u \: du = \frac{1}{2} \sin u + \mathrm{C} = \frac{1}{2} \sin 2x + \mathrm{C} = \sin x \cos x + \mathrm{C}[/tex]

    Then, we get

    [tex]\frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} \: d\theta + \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} \cos 2\theta \: d\theta = \left. \frac{1}{2} \theta + \frac{1}{2} \sin \theta \cos \theta \right] _{\sec \sqrt{2}} ^{\sec 2}[/tex]

    which is wrong.

    Any help is highly appreciated.
     
  2. jcsd
  3. May 12, 2005 #2

    dextercioby

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    Science Advisor
    Homework Helper

    Of course,the limits of integration are incorrect.

    [tex] t=2\Rightarrow \theta=\mbox{arcsec} \ 2 [/tex]

    [tex] t=\sqrt{2} \Rightarrow \theta=\mbox{arcsec} \ \sqrt{2} [/tex]

    Daniel.
     
  4. May 12, 2005 #3
    Oops! Thanks for pointing it out.

    [tex]\left. \frac{1}{2} \theta + \frac{1}{2} \sin \theta \cos \theta \right] _{\mbox{arcsec } \sqrt{2}} ^{\mbox{arcsec } 2} = \frac{1}{24}\left( -6 + 3\sqrt{3} + \pi \right)[/tex]
     
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