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Math Help - Trigonometric substitution

  1. #1
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    Trigonometric substitution

    Use the substitution x = 2tan\theta, -\pi/2 < \theta < \pi/2 to evaluate:

    \int \frac{1}{x^2 \sqrt{x^2 + 4}} dx

    Attempt:

    = \int \frac{1}{4tan^2\theta\sqrt{4tan^2\theta+ 4} } dx

    x = 2tan\theta
    dx = 2sec^2\theta d\theta

    = \int \frac {1}{(4tan^2\theta)|2sec\theta|}2sec^2\theta d\theta
    = \int \frac {sec\theta}{4tan^2\theta} d\theta
    = \frac{1}{4}\int\frac{sec\theta}{tan^2\theta}

    I dont know what to do from here.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by SyNtHeSiS View Post
    Use the substitution x = 2tan\theta, -\pi/2 < \theta < \pi/2 to evaluate:

    \int \frac{1}{x^2 \sqrt{x^2 + 4}} dx

    Attempt:

    = \int \frac{1}{4tan^2\theta\sqrt{4tan^2\theta+ 4} } dx

    x = 2tan\theta
    dx = 2sec^2\theta d\theta

    = \int \frac {1}{(4tan^2\theta)|2sec\theta|}2sec^2\theta d\theta
    = \int \frac {sec\theta}{4tan^2\theta} d\theta
    = \frac{1}{4}\int\frac{sec\theta}{tan^2\theta}

    I dont know what to do from here.
    = \frac{1}{4}\int\frac{\cos\theta d\theta}{sin^2\theta}

    Then use the substitution u = \sin \theta
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