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The following integral may seem easy to evaluate ...

$$ \int_{0}^{\Large\frac{\pi}{2}} \arctan \left(2 \tan^2 x\right) \mathrm{d}x = \pi \arctan \left( \frac{1}{2} \right). $$

Could you prove it?

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  • $\begingroup$ Have you tried the variable $u=2.tan^2(x)$ ? $\endgroup$ – D.L. Jul 22 '14 at 9:59
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    $\begingroup$ Are you challenging people to prove it? If so, do you have your own solution? You can always post it using >! before to hide it. $\endgroup$ – user37238 Jul 22 '14 at 10:15
  • $\begingroup$ @user37238 Thanks! I did not know this functionality! Where can I have more explanations about it? Yes, you have to see this integral as a small challenge. $\endgroup$ – Olivier Oloa Jul 22 '14 at 11:06
  • $\begingroup$ Have you tried compuer algebra systems? $\endgroup$ – Mhenni Benghorbal Jul 22 '14 at 11:08
  • $\begingroup$ @Mhenni Benghorbal I didn't test it, but you are right, Mathematica finds it! >$\,\frac{\pi}{2} \text{ArcTan}\left[\frac{4}{3}\right]$ $\endgroup$ – Olivier Oloa Jul 22 '14 at 11:12
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My answer is different from that you gave. Let $$ I(a)=\int_0^{\frac{\pi}{2}}\arctan(a\tan^2x)dx. $$ Than $I(0)=0$ and \begin{eqnarray} I'(a)&=&\int_0^{\frac{\pi}{2}}\frac{\tan^2x}{1+a^2\tan^4x}dx\\ &=&\int_0^\infty\frac{u^2}{(1+u^2)(1+a^2u^4)}du\\ &=&\frac{1}{1+a^2}\int_0^\infty\left(-\frac{1}{1+u^2}+\frac{1+a^2u^2}{1+a^2u^4}\right)du\\ &=&\frac{1}{1+a^2}\left(-\int_0^\infty\frac{1}{1+u^2}du+\int_0^\infty\frac{1}{1+a^2u^4}du+\int_0^\infty\frac{a^2u^2}{1+a^2u^4}\right)du\\ &=&\frac{1}{1+a^2}\left(-\frac{\pi}{2}+\frac{\pi}{2\sqrt{2}\sqrt{a}}+\frac{\sqrt{a}\pi}{2\sqrt{2}}\right) \end{eqnarray} and hence $$ I=\int_0^2I'(a)da=\pi\arctan(1+\sqrt{2a})\Big]_0^2=-\frac{\pi^2}{4}+\pi\arctan 3=\frac{\pi}{2}\arctan\frac{4}{3}.$$

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  • $\begingroup$ Thanks. Our $I$ are not the same. What is your final result for your $I$? $\endgroup$ – Olivier Oloa Jul 22 '14 at 14:34
  • $\begingroup$ why you can take the differentiation inside the integral? Can you justify it? $\endgroup$ – student Jul 22 '14 at 15:09
  • $\begingroup$ @OlivierOloa,I fixed the problem. $\endgroup$ – xpaul Jul 22 '14 at 15:41
  • $\begingroup$ @QinfengLi, I just revised the answer. I made a mistake before. $\endgroup$ – xpaul Jul 22 '14 at 15:42
  • $\begingroup$ @xpaul Please, I don't understand your antepenultimate equality. $\endgroup$ – Olivier Oloa Jul 22 '14 at 15:47
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Here is a result avoiding differentiation with respect to a parameter.

Set$$ I(\alpha):= \int_{0}^{\Large\frac{\pi}{2}} \arctan \left(\frac{2\alpha \:\sin^2 x}{\alpha^2-1+\cos^2 x}\right)\: \mathrm{d}x, \quad \alpha>0. $$ Then $$ I(\alpha)= \pi \arctan \left(\frac{1}{2\alpha}\right) \quad ({\star}) $$

With $ \alpha:=1$, we get $$ \int_{0}^{\Large\frac{\pi}{2}} \arctan \left(2 \tan^2 x\right) \mathrm{d}x = \pi \arctan \left( \frac{1}{2} \right). $$ To obtain $({\star})$ use the standard evaluation extended to complex numbers

$$ \int_{0}^{\Large\frac{\pi}{2}} \log \left(1+ t \sin^2 x\right) \mathrm{d}x = \pi \log \left( \frac{1+\sqrt{1+t}}{2} \right) $$

and observe that $$ \arctan (z) = \frac{i}{2} \left(\log (1-i z)-\log (1+i z)\right), \quad\Re z \neq 0. $$

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  • $\begingroup$ I don't understand your last two equalities. Could you please show some more details? Cause at first glance I don't think the last two equalities are easier to be obtained than the original question. $\endgroup$ – student Jul 22 '14 at 15:12

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