4
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ I asked this question on I&S... Let me find the answer from MSE. $\endgroup$ Apr 25, 2017 at 8:20

1 Answer 1

9
$\begingroup$

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

$\endgroup$
4
  • 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
    Apr 25, 2017 at 17:14
  • $\begingroup$ @FDP, could you provide more information regarding how you derived that relationship? $\endgroup$ Jul 21, 2017 at 23:29
  • $\begingroup$ Consider the change of variable $y=\sqrt{\frac{x^2}{1-x^2}}$ $\endgroup$
    – FDP
    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
    Jul 29, 2017 at 0:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.