integral_04 (Try using beta function)

$\begin{aligned} I & = \int_0^\frac \pi 2 \sin^2 t \ln (\sin t) \ dt \\ & = \int_0^\frac \pi 2 \frac 12 (1-\cos 2 t) \ln (\sin t) \ dt & \small \color{#3D99F6} \text{Using integration by parts} \\ & = \frac 12 \left( t - \frac {\sin 2t}2 \right) \ln (\sin t) \bigg|_0^\frac \pi 2 - {\color{#3D99F6} \frac 12 \int_0^\frac \pi 2 t\cot t \ dt }+ \frac 14 \int_0^\frac \pi 2 \sin 2t \cot t \ dt & \small \color{#3D99F6} \text{By integration by parts again} \\ & = 0 - {\color{#3D99F6} \frac 12 t \ln (\sin t) \bigg|_0^\frac \pi 2 + \frac 12 \int_0^\frac \pi 2 \ln (\sin t) \ dt} + \frac 12 \int_0^\frac \pi 2 \cos^2 t \ dt \\ & = 0 - {\color{#3D99F6} 0 - \frac \pi 4 \ln 2} + \frac 14 \int_0^\frac \pi 2 (1+\cos 2 t) \ dt & \small \color{#3D99F6} \text{See note: } \int_0^\frac \pi 2 \ln(\sin x) \ dx = -\frac \pi 2 \ln 2 \\ & = -\frac \pi 4 \ln 2 + \frac 14 \left[ t + \frac {\sin 2t}2 \right]_0^\frac \pi 2 \\ & = -\frac \pi 4 \ln 2 + \frac \pi 8 \\ & = \frac \pi 8 (1- \ln 4) \end{aligned}$

Therefore, $a+b+c+d = 1+8+1+4 = \boxed{14}$ .

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Note:

$\begin{aligned} J & = \int_0^\frac \pi 2 \ln (\sin x)\ dx & \small \color{#3D99F6} \text{By identity: }\int_a^b f(x)\ dx = \int_a^b f(a+b-x)\ dx \\ & = \frac 12 \int_0^\frac \pi 2 (\ln (\sin x) + \ln (\cos x))\ dx \\ & = \frac 12 \int_0^\frac \pi 2 \ln (\sin x \cos x)\ dx \\ & = \frac 12 \int_0^\frac \pi 2 \ln \left(\frac {\sin 2x}2 \right)\ dx \\ & = {\color{#3D99F6} \frac 12 \int_0^\frac \pi 2 \ln (\sin 2x)\ dx} - \frac 12 \int_0^\frac \pi 2 \ln 2 \ dx & \small \color{#3D99F6} \text{Let } u = 2x \implies du = 2\ dx \\ & = {\color{#3D99F6} \frac 14 \int_0^\pi \ln (\sin u)\ du} - \frac \pi 4 \ln 2 & \small \color{#3D99F6} \text{Since } \ln (\sin u) \text{ is symmetrical about }u = \frac \pi 2 \\ & = {\color{#3D99F6} \frac 12 \int_0^\frac \pi 2 \ln (\sin u)\ du} - \frac \pi 4 \ln 2 & \small \color{#3D99F6} \text{Note that } \int_0^\frac \pi 2 \ln (\sin u)\ du = J \\ & = {\color{#3D99F6} \frac 12 J} - \frac \pi 4 \ln 2 \\ & = - \frac \pi 2 \ln 2 \end{aligned}$

$\begin{aligned} & \beta(x,y)=2\int_{0}^{\frac{\pi}{2}}\sin^{2x-1}(t)\cos^{2y-1}(t)dt\\ & \frac{d}{dx}\beta(x,y)=2\frac{d}{dx}\int_{0}^{\frac{\pi}{2}}\sin^{2x-1}(t)\cos^{2y-1}(t)dt\\ & \beta(x,y)\left(\psi(x)-\psi(x+y)\right)=4\int_{0}^{\frac{\pi}{2}}\sin^{2x-1}(t)\cos^{2y-1}(t)\ln(\sin(t))dt & & \color{#D61F06}\frac{d}{dx}B(x,y)=B(x,y)\left(\psi(x)-\psi(x+y)\right)\\ & \beta(\frac{3}{2},\frac{1}{2})\left(\psi(\frac{3}{2})-\psi(2)\right)=4\int_{0}^{\frac{\pi}{2}}\sin^2(t)\ln(\sin(t))dt & & \color{#D61F06}x=\frac{3}{2},y=\frac{1}{2}\\ & \frac{\pi}{2}\left(\left(2-2\ln(2)-\gamma\right)-\left(1-\gamma\right)\right)=4I & & \color{#D61F06}\beta(\frac{3}{2},\frac{1}{2})=\frac{\pi}{2},\psi(\frac{3}{2})=2-2\ln(2)-\gamma,\psi(2)=1-\gamma\\ & I=\frac{\pi}{8}\left(1-2\ln(2)\right)=\frac{\pi}{8}\left(1-\ln(4)\right)\\ & a=1,b=8,c=1,d=4 \\ & \color{#3D99F6} \text{Note:}\\ & \color{#3D99F6} \beta(\frac{3}{2},\frac{1}{2})=\frac{\Gamma(\frac{3}{2})\Gamma(\frac{1}{2})}{\Gamma(\frac{3}{2}+\frac{1}{2})}=\frac{\frac{1}{2}\Gamma(\frac{1}{2})\Gamma(\frac{1}{2})}{\Gamma(2)}=\frac{\pi}{2}//\Gamma(\frac{1}{2})=\sqrt\pi//\Gamma(n+1)=n\Gamma(n)\\ & \color{#3D99F6} \psi(\frac{3}{2})=\psi(1+\frac{1}{2})=2-\gamma-2\ln(2)//\psi(n+\frac{1}{2})=-\gamma-2\ln(2)-\sum_{k=1}^n\frac{1}{2k-1}=\\ & \color{#3D99F6} \psi(2)=1+\psi(1)//\psi(n+1)=\frac{1}{n}+\psi(n)//\psi(1)=-\gamma \end{aligned}\\$

$\int \sin ^2(t) \log (\sin (t)) \, dt \Rightarrow \frac{1}{8} \left(2 i \text{Li}_2\left(e^{2 i t}\right)+2 t \left(i t-2 \log \left(1-e^{2 i t}\right)+2 \log (\sin (t))+1\right)+\sin (2 t) (1-2 \log (\sin (t)))\right)$

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$\frac{1}{8} \left(2 i \text{Li}_2\left(e^{2 \mathbb{i} t}\right)+2 t \left(\mathbb{i} t-2 \log \left(1-e^{2 \mathbb{i} t}\right)+2 \log (\sin (t))+1\right)+\sin (2 t) (1-2 \log (\sin (t)))\right)$ $\Rightarrow$ $(\frac{1}{8} \left(\pi \left(1+\frac{\mathbb{i} \pi }{2}-2 \log (2)\right)-\frac{1}{6} \left(\mathbb{i} \pi ^2\right)\right))-(\frac{\mathbb{i} \pi ^2}{24})$ $\Rightarrow$ $\frac{1}{8} \pi ^1 (1-\log (4))$