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I was going through the the top votes questions and I saw this quite of interesting post posted by @Sangchul Lee. I was just messing around with it and found a slightest Variation of Ahmed's integral.

$$\int_{0}^{1}{\arctan(x\sqrt{x^2+2})\over \sqrt{x^2+2}}\cdot{dx\over x^2+1}=\left({\pi\over 6}\right)^2\tag1$$

May be substitution might work

$u=x(x^2+2)^{1/2}$ the $du=x(x^2+2)^{-1/2}+(x^2+2)^{1/2}dx$

$$\int_{\sqrt{2}}^{4\sqrt{3}\over3}{x^2\arctan{u}\over x^3+u^2}\cdot{x^2\over u^2-x^2}du$$

More difficult then before. I can't remove the x variable from the integral.

Any suggestion on how to prove $(1)$?

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Using your approach $$u=x(x^2+2)^{1/2}\implies x=\sqrt{\sqrt{u^2+1}-1}\implies dx=\frac{u}{2 \sqrt{u^2+1} \sqrt{\sqrt{u^2+1}-1}}\,du$$ and, after simplifications $$\int{\arctan(x\sqrt{x^2+2})\over \sqrt{x^2+2}}\cdot{dx\over x^2+1}=\int\frac{\tan ^{-1}(u)}{2 u^2+2}\,du=\frac{1}{4} \tan ^{-1}(u)^2$$

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  • $\begingroup$ Very nice, Claude. The OP made a mistake while calculating $\;\frac{dx}{du}\;$ ...+1 $\endgroup$ – DonAntonio Jan 8 '17 at 8:59
  • $\begingroup$ How did you make x the subject in term of u @Claude? I don't really get it! $\endgroup$ – user339807 Jan 8 '17 at 16:01
  • $\begingroup$ $u^2=x^2(x^2+2)\implies x^4+2x^2-u^2=0$ which is a quadratic in $x^2$. $\endgroup$ – Claude Leibovici Jan 8 '17 at 16:52
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Substitute : $u=\arctan\left(x\sqrt{x^2+2}\right)$

You will get : $$\dfrac{\mathrm{d}u}{\mathrm{d}x}=\dfrac{\sqrt{x^2+2}+\frac{x^2}{\sqrt{x^2+2}}}{x^2\left(x^2+2\right)+1}$$

Thus : $$\int_{0}^{1}{\arctan(x\sqrt{x^2+2})\over \sqrt{x^2+2}}\cdot{dx\over x^2+1}=\int_{0}^{\arctan(\sqrt{3})}{u\over 2}\cdot{du}=\left[\frac{u^2}{4}\right]^{\frac{\pi}{3}}_0=\left({\pi\over 6}\right)^2$$

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  • $\begingroup$ Very nice answer. +1 $\endgroup$ – DonAntonio Jan 8 '17 at 8:56

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