Integral and Gamma Function (4)

$\begin{aligned} I&=\int_0^1(\ln\Gamma(x))\cos^2(\pi x)\ dx&\small\color{#3D99F6}\text{Using }\int_a^bf(x)dx=\int_a^bf(a+b-x)\ dx\\ &=\frac12\int_0^1(\ln\Gamma(x)+\ln\Gamma(1-x))\cos^2(\pi x)\ dx\\ &=\frac12\int_0^1(\ln\Gamma(x)\Gamma(1-x))\cos^2(\pi x)\ dx&\small\color{#3D99F6}\text{Using Euler's reflection formula}\\ &=\frac12\int_0^1\left(\ln\frac\pi{\sin(\pi x)}\right)\cos^2(\pi x)\ dx\\ &=\frac{\ln\pi}2\int_0^1\cos^2(\pi x)\ dx-\color{#3D99F6}\frac12\int_0^1(\ln\sin(\pi x))\cos^2(\pi x)\ dx&\small\color{#3D99F6}\text{Let }u=\pi x\implies du=\pi\ dx\\ &=\frac{\ln\pi}4\int_0^1(1+\cos(2\pi x))\ dx-\color{#3D99F6}\frac1{2\pi}\int_0^\pi(\ln\sin u)\cos^2 u\ du&\small\color{#3D99F6}\text{See note}\\ &=\frac{\ln\pi}4-\color{#3D99F6}\frac 1{2\pi}\left(-\frac {\pi\ln 2}2-\frac \pi4\right)\\ &=\frac {\ln(2\pi)}4+\frac 18 \end{aligned}$

Therefore, $A+B+C=4+1+8=\boxed{13}$ .

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Note:

$\begin{aligned} I_1&=\int_0^\pi(\ln\sin u)\cos^2 u\ du\\ &=\frac 12\int_0^\pi(\ln\sin u)(1+\cos (2u))\ du\\ &=\color{#3D99F6}\frac 12\int_0^\pi(\ln\sin u)\ du\color{#333333}+\frac 14\int_0^\pi\ln\sin u\ d\sin(2u)&\small\color{#3D99F6}\sin u\text{ is symmetrical about }u=\frac \pi 2\\ &=\color{#3D99F6}\int_0^{\pi /2}(\ln\sin u)\ du\color{#333333}+\frac 14\left((\ln\sin u)(\sin 2u)\bigg|_0^\pi-\int_0^\pi\frac {\cos u}{\sin u}\sin (2u)\ du\right)&\small\color{#3D99F6}\text{See the problem }Using\ symmetry\ (advanced)\\ &=\color{#3D99F6}-\frac {\pi\ln 2}2\color{#333333}+\frac 14\left(0-\int_0^\pi2\cos^2u\ du\right)\\ &=-\frac {\pi\ln 2}2-\frac 14\int_0^\pi(1+\cos (2u))\ du\\ &=-\frac {\pi\ln 2}2-\frac \pi4 \end{aligned}$

Here is Using symmetry (advanced) .