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I had to calculate $$I= \int_{0}^{1}\frac{1}{1+x^2}\ln\bigg[\frac{x^2+x+1}{x^2-x+1}\bigg]\,\mathrm dx$$

from my Previous Question [1](which is now solved) but I wanted to have another solution ,So I proceed like this.

$$I=\int_{0}^{1}\frac{\ln(x^2+x+1)}{x^2+1}\,\mathrm dx-\int_{0}^{1}\frac{\ln(x^2-x+1)}{x^2+1}\,\mathrm dx$$

Let $x=\tan t \implies \mathrm dx=\sec^2 t \,\mathrm dt$

$$\implies I=\int_{0}^{\frac{\pi}{4}}\ln(\tan^2 x+\tan x+1)\,\mathrm dx-\int_{0}^{ \frac{\pi}{4}}\ln(\tan^2 x-\tan x+1)\,\mathrm dx$$

$$2(\tan^2 x+\tan x+1)=(2+\sin 2x)(1+\tan^2 x)$$

$$\implies I=\int_{0}^{ \frac{\pi}{4} }\Big[\ln(2+\sin 2x)+\ln(1+\tan^2 x)-\ln(2)\Big]\,\mathrm dx-\int_{0}^{ \frac{\pi}{4} }\Big[\ln(2-\sin 2x)+\ln(1+\tan^2 x)-\ln(2)\Big]\,\mathrm dx$$

$$\implies I=\frac12 \Bigg[\int_{0}^{ \frac{\pi}{2} }\ln(2+\sin x)\,\mathrm dx- \int_{0}^{ \frac{\pi}{2} }\ln(2-\sin x)\,\mathrm dx\Bigg]$$

We can group together the two log terms in the integral but then this Question will become same as my previous question[1], I want to calculate both of the integrals separately.

Bernard
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User 1207
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    You have typed sin\space x, so that the reader sees $sin\space x.$ But if you type \sin x, then the viewer sees $\sin x,$ and that is standard usage. With \sin x and \sin(x) you see $\sin x$ and $\sin(x),$ with more horizontal space to the right of $\sin$ in the former than in the latter, i.e. the spacing is context -dependent (and the same applies to the left of $\sin$). Similarly with $\tan x$ and $\log x$: Use \tan and \ln. $\qquad$ – Michael Hardy Jun 30 '21 at 17:17
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    $\int_0^{\frac{\pi}{2}}\ln(4-\sin^2x)dx= \Bigl(\int_{0}^{ \frac{\pi}{2} }\ln(2+\sin x) dx+ \int_{0}^{ \frac{\pi}{2} }\ln(2-\sin x)dx\Bigr)$ https://artofproblemsolving.com/community/c7h567792p3328154. And difference you already know – Svyatoslav Jun 30 '21 at 18:05
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    @svyatoslav Nice trick.This trick reminds me of the this Question $2^{nd}$ answer https://math.stackexchange.com/questions/199094/evaluating-int-0-large-frac-pi4-log-left-cos-x-right-mathrmdx – User 1207 Jun 30 '21 at 18:31

1 Answers1

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Let $J(a) = \int_0^{\frac\pi2}\ln(1+\sin a\sin x)dx$. Then $$J’(a) =\int_0^{\frac\pi2}\frac{\cos a\sin x}{1+\sin a\sin x}dx =a\csc a-\frac\pi2\tan \frac a2 $$ and \begin{align} \int_{0}^{\frac{\pi}{2}}\ln(2+\sin x)& dx =\frac\pi2 \ln 2+J(\frac\pi6) =\frac\pi2 \ln 2+ \int^{\frac{\pi}6}_{0} \overset{a=2t}{ J’(a)da }\\ =&\frac\pi2 \ln 2 -\pi\int^{\frac{\pi}{12}}_{0}\tan t \>dt + 4\int^{\frac{\pi}{12}}_{0}\overset{IBP}{ t \csc 2t\>dt }\\ =&\frac\pi2 \ln 2+\pi \ln (\cos t)\bigg|_0^{\frac\pi{12}} -2t\ln(\cot t)\bigg|_0^{\frac\pi{12}} +2 \int_{0} ^{\frac\pi{12}} {\ln(\cot t) dt}\\ =&\frac{\pi}{3} \ln(2+\sqrt3)-\frac\pi2\ln2+\frac43 G \end{align} where $\int^{ \frac\pi{12}}_{0} \ln(\cot t)dt= \frac23G$. Similarly \begin{align} & \int_{0}^{\frac{\pi}{2}}\ln(2-\sin x)dx =\frac\pi2 \ln 2+J(-\frac\pi6) =\frac{2\pi}{3} \ln(2+\sqrt3)-\frac\pi2\ln2-\frac43 G \end{align}

Quanto
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  • (+1) WOW, I didn't thought of that even after reading your answer on my previous question. – User 1207 Jul 01 '21 at 03:20
  • If I am not wrong, there is a problem with $J'(a)$. Have a look at https://math.stackexchange.com/questions/4187503/for-what-interval-of-a-is-j-a-int-0-frac-pi2-frac-cos-a-sin-x1-sin/4187525#4187525 . This would be very surprising taking into account your talent with (at least) integrals. Please, let me know. Thanks & cheers :-) – Claude Leibovici Jul 01 '21 at 06:13
  • @ClaudeLeibovici - Implicitly, $a\in (-\pi/2,\pi/2)$ per the definition of $J(a)$, for which $J’(a) =a\csc a-\frac\pi2\tan \frac a2$ is valid and verifiable numerically. – Quanto Jul 01 '21 at 12:07