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Is there somone who can show me how do I evaluate this integral :$$\int \frac{1}{ \sqrt{1-x^2} } \arctan\left(\frac{\sqrt{1-x^2}}{2}\right)dx$$

Note: I took : $x=\cos t$ , but it didn't work

Thank you for any help .

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  • $\begingroup$ According to Wolfie, it involves some pretty complicated stuff... wolframalpha.com/input/?i=int+1%2Fsqrt%281-x%5E2%29+*+arctan%28sqrt%281-x%5E2%29%2F2%29 $\endgroup$ Jul 13, 2015 at 14:23
  • $\begingroup$ do you think that i didn't get attention for that ? $\endgroup$ Jul 13, 2015 at 14:24
  • $\begingroup$ I don't know, but good luck finding a closed form for that integral. $\endgroup$ Jul 13, 2015 at 14:26
  • $\begingroup$ @CauchytheDog I've deleted my comment. Earlier, your link was not appearing in its entirety. But now I see the entire URL. $\endgroup$ Jul 13, 2015 at 15:09
  • $\begingroup$ @dog , if you are really a dog give me a closed form of the above form, your link not work $\endgroup$ Jul 13, 2015 at 15:12

2 Answers 2

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We have:

$$\begin{eqnarray*} \int_{0}^{u}\arctan\left(\frac{\cos\theta}{2}\right)\,d\theta &=& \int_{0}^{u}\int_{0}^{\frac{\cos\theta}{2}}\frac{1}{1+t^2}\,dt\,d\theta\\&=&\int_{0}^{u}2\cos\theta\int_{0}^{1}\frac{1}{4+z^2\cos^2\theta}\,dz\,d\theta\\&=&\int_{0}^{\tan u}\frac{1}{\sqrt{1+v^2}}\int_{0}^{1/2}\frac{1}{1+v^2+z^2}\,dz\,dv\\&=&\int_{0}^{\text{arcsinh}\tan u}\int_{0}^{1/2}\frac{1}{z^2+\cosh^2\rho}\,dz\,d\rho\\&=&\int_{0}^{1/2}\text{arctanh}\left(\frac{z\sin u}{\sqrt{1+z^2}}\right)\frac{dz}{z\sqrt{1+z^2}}\\&=&\int_{2}^{+\infty}\text{arctanh}\left(\frac{\sin u}{\sqrt{1+w^2}}\right)\frac{dw}{\sqrt{1+w^2}}\\&=&\int_{\text{arcsinh} 2}^{+\infty}\text{arctanh}\left(\frac{\sin u}{\cosh \eta}\right)\,d\eta\\&=&\color{red}{\int_{2+\sqrt{5}}^{+\infty}\text{arctanh}\left(\frac{2\tau}{1+\tau^2}\sin u\right)\frac{d\tau}{\tau}}\end{eqnarray*}$$ hence the answer is given by a combination of dilogarithms, since $\text{arctanh}(x)=\frac{1}{2}\left(\log(1+x)-\log(1-x)\right)$ and:

$$ \int\log(1+\tau^2)\frac{d\tau}{\tau}=-\frac{1}{2}\text{Li}_2(-\tau^2),\qquad \int\log(1+\alpha\tau)\frac{d\tau}{\tau}= -\text{Li}_2(-\alpha\tau).$$

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For the integral \begin{align} I = \int \frac{1}{ \sqrt{1-x^2} } \, \arctan\left(\frac{\sqrt{1-x^2}}{2}\right) \, dx \end{align} let $u^{2} = 1 - x^{2}$ to obtain \begin{align} I = - \, \int \tan^{-1}\left(\frac{u}{2}\right) \, \frac{du}{\sqrt{1-u^{2}}}. \end{align} Using $2 \tan^{-1}(x) = \ln(1 - ix) - \ln(1+ix)$ then after the transformation $2 t = i u$ the integral becomes \begin{align} I = i \, \int \ln\left( \frac{1-t}{1+t} \right) \, \frac{dt}{\sqrt{1 + 4 t^{2}}}. \end{align} Making use of integration by parts where \begin{align} u = \ln\left(\frac{1-x}{1+x}\right) \hspace{10mm} dv = \frac{dt}{\sqrt{1 + 4 t^{2}}} \end{align} to obtain \begin{align} -i \, I = \frac{1}{2} \, \ln\left(\frac{1-t}{1+t}\right) \, \sinh^{-1}(2t) - \int \frac{\sinh^{-1}(2t) \, dt}{t^{2} -1}. \end{align} From here use Wolfram Alpha to reduce the necessary terms.

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  • $\begingroup$ Well done, but the main issue is that WA rarely gives a combination of dilogarithms in its most compact form. $\endgroup$ Jul 13, 2015 at 17:07
  • $\begingroup$ @JackD'Aurizio It is true the WA tends not to the most efficient for some things $\endgroup$
    – Leucippus
    Jul 13, 2015 at 17:25

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