Calculate $\int_{0}^{\pi/2}\arccos\frac{\cos x}{1+2\cos x}\,\mathrm dx.$

Asked Nov 30 '22 at 23:57

Active Dec 01 '22 at 00:04

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I've stumbled across this definite integral and was unable to find a closed form, can anyone help me out?

This integral $I$ is defined as follows: $$ I \equiv \int_{0}^{\pi/2}\arccos\left(\frac{\cos(x)}{1+2\cos(x)}\right)dx $$

To solve this problem, I've tried to incorporate the double angle identity $\cos(2\theta) = 2\cos^2(\theta)-1$ as follows:

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

edited Dec 01 '22 at 00:04

asked Nov 30 '22 at 23:57

Moon月

To get rid of the $\arccos$ we could integrate by parts with $dv=dx$. – aschepler Dec 01 '22 at 00:11
FWIW, an approximate numerical answer is 2.05617. – Dan Dec 01 '22 at 01:26
@aschepler I'll try that, thanks. – Moon月 Dec 01 '22 at 02:01
1

@Dan the given solution is $\frac{5\pi^2}{24}$, I'm just not sure how to find it. – Moon月 Dec 01 '22 at 02:02
A solution can be found here. – Accelerator Dec 01 '22 at 03:08
Okay, thank you! – Moon月 Dec 01 '22 at 04:36
2

For a (hopefully) simpler solution, your integral can be rewritten as $$I_1=\int_0^{\frac{\pi}{2}}\sec^{-1}(\sec x + 2)dx$$ with inverse $$I_2=\int_{\sec^{-1}(3)}^{\frac{\pi}{2}}\sec^{-1}(\sec y - 2)dy$$ Pictorally, the two integrals add up to the area of the square $$I_1+I_2 = \frac{\pi^2}{4}$$ If you can prove that $I_1 = 5 I_2$, the proof is complete. – Ninad Munshi Dec 01 '22 at 05:05
That is a creative method, I will definitely try to prove that. Thanks! – Moon月 Dec 01 '22 at 05:15
see https://math.stackexchange.com/questions/4037652/solve-int-0-pi-2-arccos-left-frac-cosx12-cosx-right-mathrm – FDP Dec 01 '22 at 09:59

Calculate $\int_{0}^{\pi/2}\arccos\frac{\cos x}{1+2\cos x}\,\mathrm dx.$

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