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Math Help - Help with trigonometric integration

  1. #1
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    Help with trigonometric integration

    I have to prove that the integral of 1/(u^2+2) is tan^(-1)(u/sqrt(2)))/sqrt(2)

    So far, I've been able to show that 1/(u^2+2) = 1/4(
    Θ +sin(Θ)cost(Θ)) where Θ=tan^(-1)(u/sqrt(2))

    Any help would be appreciated


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  2. #2
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    Re: Help with trigonometric integration

    Quote Originally Posted by Giestforlife View Post
    I have to prove that the integral of 1/(u^2+2) is tan^(-1)(u/sqrt(2)))/sqrt(2)

    So far, I've been able to show that 1/(u^2+2) = 1/4(
    Θ +sin(Θ)cost(Θ)) where Θ=tan^(-1)(u/sqrt(2))

    Any help would be appreciated



    Make the substitution \displaystyle \begin{align*} u = \sqrt{2}\tan{\theta} \implies du = \sqrt{2}\sec^2{\theta}\,d\theta \end{align*} and the integral becomes...

    \displaystyle \begin{align*} \int{\frac{1}{u^2 + 2}\,du} &= \int{\frac{1}{\left(\sqrt{2}\tan{\theta}\right)^2 + 2}\,\sqrt{2}\sec^2{\theta}\,d\theta} \\ &= \sqrt{2}\int{\frac{\sec^2{\theta}}{2\tan^2{\theta} + 2}\,d\theta} \\ &= \frac{\sqrt{2}}{2}\int{\frac{\sec^2{\theta}}{1 + \tan^2{\theta}}\,d\theta} \\ &= \frac{\sqrt{2}}{2}\int{\frac{\sec^2{\theta}}{\sec^  2{\theta}}\,d\theta} \\ &= \frac{\sqrt{2}}{2}\int{1\,d\theta} \\ &= \frac{\sqrt{2}}{2}\theta + C \\ &= \frac{\sqrt{2}}{2}\arctan{\left(\frac{\sqrt{2}}{2}  \,u\right)} + C \end{align*}
    Thanks from Giestforlife
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