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Math Help - definite integral #3

  1. #1
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    definite integral #3

     \displaystyle \int_{0}^{1} ( \arctan x )^{2} \ dx
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  2. #2
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    Starting with integration by parts (twice), we have:

    \displaystyle \begin{aligned}  \int_{0}^{1}\left(\tan^{-1}(x)\right)^2\;{dx} &= x(\tan^{-1}{x})^2\bigg|_{ 0}^{1}-\int_{0}^{1} \frac{2x\tan^{-1}{x}}{1+x^2}\;{dx} \\& = \frac{\pi^2}{16}- \int_{0}^{1} \left[\ln(x^2+1)\right]'\tan^{-1}{x}\;{dx} \\&= \frac{\pi^2}{16}-\ln(x^2+1)\tan^{-1}{x}\bigg|_{0}^{1}+\int_{0}^{1} \frac{\ln\left(x^2+1\right)}{x^2+1}\;{dx} \\& = \frac{\pi^2}{16}-\frac{\pi}{4}\ln(2)+\int_{0}^{1} \frac{\ln\left(x^2+1\right)}{x^2+1}\;{dx} \\& = \frac{\pi^2}{16}-\frac{\pi}{4}\ln(2)+I_{1}.\end{aligned}

    Where letting u = \arctan{x}  \Rightarrow \frac{du}{dx} = \frac{1}{1+x^2} \Rightarrow dx = (1+x^2)\;{du} gives:

    \displaystyle \begin{aligned} I_{1}  & = \int_{0}^{1} \frac{\ln\left(x^2+1\right)}{x^2+1}\;{dx} = \int_{0}^{\frac{\pi}{4}}\frac{\ln(\tan^2{u}+1)(1+x  ^2)}{1+x^2}\;{du} \\& = \int_{0}^{\frac{\pi}{4}}\ln(\sec^2{u})\;{du} = -2\int_{0}^{\frac{\pi}{4}}\ln(\cos{u})\;{du} \\& = -2\int_{0}^{\frac{\pi}{4}}\ln\left(2\cos{u}}\right) \;{du}+2\int_{0}^{\frac{\pi}{4}}\ln(2)\;{du} \\& = -K+\frac{\pi}{2}\ln(2). \end{aligned}

    Where K is Catalan's constant.

    Thus \displaystyle \int_{0}^{1} ( \arctan x )^{2} \;{dx} = \frac{\pi^2}{16}+\frac{\pi}{4}\ln(2)-K.
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