# Integral $\int_{0}^{\pi/2} \arctan \left(2\tan^2 x\right) \mathrm{d}x$

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?

• Have you tried the variable $u=2.tan^2(x)$ ? – D.L. Jul 22 '14 at 9:59
• 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. – user37238 Jul 22 '14 at 10:15
• @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. – Olivier Oloa Jul 22 '14 at 11:06
• Have you tried compuer algebra systems? – Mhenni Benghorbal Jul 22 '14 at 11:08
• @Mhenni Benghorbal I didn't test it, but you are right, Mathematica finds it! >$\,\frac{\pi}{2} \text{ArcTan}\left[\frac{4}{3}\right]$ – Olivier Oloa Jul 22 '14 at 11:12

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 $$I=\int_0^2I'(a)da=\pi\arctan(1+\sqrt{2a})\Big]_0^2=-\frac{\pi^2}{4}+\pi\arctan 3=\frac{\pi}{2}\arctan\frac{4}{3}.$$

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

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

• 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. – student Jul 22 '14 at 15:12