TITLE: Evaluate $\int_{0}^{\frac{\pi}{2}}\ln(2+\sin x) \,\mathrm dx $ and $\int_{0}^{\frac{\pi}{2}}\ln(2-\sin x) \,\mathrm dx$
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: 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}