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I stumbled across this definite integral and was unable to find a closed form.

This integral is defined as follows: $$ I \equiv \int_{0}^{\pi/2} \arccos\left(\frac{\cos \left(x\right)} {1 + 2\cos\left(x\right)}\right){\rm d}x $$ To solve this problem, I tried to incorporate the double angle identity $\cos\left(2\theta\right) = 2\cos^{2}\left(\theta\right) - 1$ as follows:

$$ \mbox{Isolate}\ \theta:\quad 2\theta = \arccos\left(2\cos^{2}\left(\theta\right) - 1\right) $$ $$ \mbox{Let}\quad \alpha = 2\cos^{2}\left(\theta\right) - 1 $$ After substituting and some manipulation $$ \theta = \arccos\left(\sqrt{\frac{1 + \alpha}{2}}\right) $$ $$ \mbox{Hence,}\quad \arccos(\alpha) = 2\arccos\left(\sqrt{\frac{1+\alpha}{2}}\right) $$ This finding enables us to rewrite $I$ as the following $$ I = \int_{0}^{\pi/2}2\arccos\left(\sqrt{\frac{1 + 3\cos\left(x\right)}{2 + 4\cos\left(x\right)}}\right){\rm d}x $$ However, I am not sure how to continue from here, so any help would be greatly appreciated.

Felix Marin
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Moon月
  • 329

2 Answers2

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Here is an efficient procedure for this challenging integral, also known as the Coxeter’s integral. Utilize the identity $\cos^{-1}a=2\cot^{-1}\sqrt{\frac{1+a}{1-a}}$, followed by the half-angle substitution $t=\tan\frac x2$

\begin{align} I =&\int_{0}^{\pi/2}\cos^{-1}\frac{\cos x}{1+2\cos x}dx\\ =&\int_{0}^{\pi/2}2\cot^{-1}\sqrt{\frac{1+3\cos x}{1+\cos x}}dx=4\int_0^1 \frac{\cot^{-1}\sqrt{2-t^2}}{1+t^2}\ \overset{ibp}{dt}\\ =&\ \frac{\pi^2}{4}-4\int_0^1 \frac{t\tan^{-1}t}{(3-t^2) \sqrt{2-t^2}}dt\>\>\>\>\>\>\>\>(2-t^2\to t^2)\\ =& \ \frac{3\pi^2}{4}-4\int_0^{\sqrt2}\frac{\tan^{-1}\sqrt{2-t^2}}{1+t^2}dt-I\\ =& \ \frac{3\pi^2}{8}-2\int_0^{\sqrt2}\frac{\tan^{-1}\sqrt{2-t^2}}{1+t^2}dt = \frac{3\pi^2}{8}-2\cdot \frac{\pi^2}{12}=\frac{5\pi^2}{24} \end{align} where \begin{align} &\int_0^{\sqrt2}\frac{\tan^{-1}\sqrt{2-t^2}}{1+t^2}dt\\ = & \int_0^{\sqrt2}\int_0^1\frac{\sqrt{2-t^2}}{(1+t^2)(1+2s^2-t^2s^2)}ds\ dt\\ =&\int_0^1 \frac{\frac\pi2}{3s^2+1}\bigg( {\sqrt3}- \frac1{\sqrt{2s^2+1}}\bigg)ds=\frac{\pi^2}{12} \end{align} Note that the above shortcut avoids traversing through the Ahmed integral, which is often done elsewhere.

Quanto
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\begin{align}J&=\int_0^{\frac{\pi}{2}}\arccos\left(\frac{\cos x}{1+2\cos x}\right)dx\\ &\overset{u=\tan\left(\frac x2\right)}=2\int_0^1\frac{\arccos\left(\frac{1-u^2}{3-u^2}\right)}{1+u^2}du\\ &\overset{\text{IBP}}=2\left[\arccos\left(\frac{1-u^2}{3-u^2}\right)\arctan u\right]_0^1-4\int_0^1\frac{u\arctan u}{(3-u^2)\sqrt{2-u^2}}du\\ &=\frac{\pi^2}{4}-4\underbrace{\int_0^1\frac{u\arctan u}{(3-u^2)\sqrt{2-u^2}}du}_{=K}\\ F(a)&=\int_0^1\frac{u\arctan(au)}{(3-u^2)\sqrt{2-u^2}}du\\ F^\prime(a)&=\int_0^1\frac{u^2}{(1+a^2u^2)(3-u^2)\sqrt{2-u^2}}du\\ &=\left[\frac{\sqrt{3}\arctan\left(\frac{u}{\sqrt{6-3u^2}}\right)}{1+3a^2}-\frac{\arctan\left(\frac{u\sqrt{1+2a^2}}{\sqrt{2-u^2}}\right)}{(1+3a^2)\sqrt{1+2a^2}}\right]_0^1\\ &=\frac{\pi}{2(1+3a^2)\sqrt{3}}-\frac{\arctan\left(\sqrt{1+2a^2}\right)}{(1+3a^2)\sqrt{1+2a^2}}\\ K&=F(1)-F(0)=\int_0^1F^\prime(a)da=\frac{\pi^2}{18}-\underbrace{\int_0^1\frac{\arctan\left(\sqrt{1+2a^2}\right)}{(1+3a^2)\sqrt{1+2a^2}}da}_{=L}\\ \end{align} Moreover $\displaystyle L=\dfrac{13\pi^2}{288}$ thus, \begin{align}K&=\frac{\pi^2}{96},\boxed{J=\frac{5\pi^2}{24}} \end{align}

FDP
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