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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- Aug 5th 2012, 10:22 AMGiestforlifeHelp 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

- Aug 5th 2012, 10:29 AMProve ItRe: Help with trigonometric integration

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

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