This question was posted on I&S

Prove the following

$$\int_0^\frac{\pi}{2} \cot^{-1}{\sqrt{1+\csc{\theta}}}\,\mathrm d\theta =\frac{\pi^2}{12}$$

The numerical value of the integral seems to agree with the answer.

Maybe someone could use that

$$1+\csc \theta = \csc(\theta) (\cos(\theta/2) + \sin(\theta/2))^2$$

I am sure there is a smart substitution or some trigonometric properties that I fail to see.

  • $\begingroup$ I asked this question on I&S... Let me find the answer from MSE. $\endgroup$ Commented Apr 25, 2017 at 8:20

1 Answer 1


The first integral is $$ I_1=\frac{\pi^2}{4}-\int_{0}^{\pi/2}\arctan\sqrt{\frac{1+\sin t}{\sin t}}\,dt=\frac{\pi^2}{4}-2\int_{0}^{\pi/4}\arctan\sqrt{\frac{1+\cos(2t)}{\cos (2t)}}\,dt$$ and $$\int_{0}^{\pi/4}\arctan\sqrt{\frac{1+\cos(2t)}{\cos (2t)}}\,dt=\int_{0}^{1}\arctan\sqrt{\frac{2}{1-u^2}}\frac{du}{1+u^2}$$ is a variant of Ahmed's integral that can be tackled through differentiation under the integral sign: it is enough to be able to integrate $\frac{\sqrt{1-u^2}}{(1+a-u^2)(1+u^2)}$.

Credit goes to Jack D'Aurizio

  • 1
    $\begingroup$ $\displaystyle \int_0^1\arctan\left(\sqrt{\dfrac{2}{1-x^2}}\right)\cdot \dfrac{1}{1+x^2}dx=\int_0^{\infty}\dfrac{\arctan\left(\sqrt{2+x^2}\right)}{(1+x^2)\sqrt{2+x^2}}dx$ $\endgroup$
    – FDP
    Commented Apr 25, 2017 at 17:14
  • $\begingroup$ @FDP, could you provide more information regarding how you derived that relationship? $\endgroup$ Commented Jul 21, 2017 at 23:29
  • $\begingroup$ Consider the change of variable $y=\sqrt{\frac{x^2}{1-x^2}}$ $\endgroup$
    – FDP
    Commented Jul 29, 2017 at 0:49
  • $\begingroup$ It's natural to try $\sqrt{\frac{2}{1-x^2}}=\sqrt{2+y^2}$. this leads to the change of variable mentioned above. $\endgroup$
    – FDP
    Commented Jul 29, 2017 at 0:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .