A calculus problem by chickpeas and tahini

$\text{Starting with } \mathfrak{I} = 2\int_{0}^{\frac{\pi}{2}} ln(sin2x)dx$

$\text{Substitute 2x=t , reduce the integral \& replace t with x and it is now : } \int_{0}^{\pi} ln(sinx)dx$

$\text{Using } \int_{0}^{2a}f(x)dx=2\int_{0}^{a}f(x)dx \text{ if f(2a-x)=f(x) we get } \frac{1}{2}\mathfrak{I} = \int_{0}^{\frac{\pi}{2}} ln(sinx)dx$

$\begin{aligned} \frac{1}{2}\mathfrak{I} = \int_{0}^{\frac{\pi}{2}} ln(sinx)dx = \int_{0}^{\frac{\pi}{2}} ln(cosx)dx\end{aligned}$

$\begin{aligned} \mathfrak{I} = \int_{0}^{\frac{\pi}{2}} (ln(sinx) + ln(cosx))dx \implies \mathfrak{I} = \int_{0}^{\frac{\pi}{2}} ln(\frac{sin2x}{2})dx = \int_{0}^{\frac{\pi}{2}} (ln(sin2x) - ln2)dx\end{aligned}$

$\mathfrak{I} = \underbrace{\color{#D61F06}{\int_{0}^{\frac{\pi}{2}} ln(sin2x)dx}} - (\frac{\pi}{2}ln2)$

$\text{The braced integral is nothing but } \frac{1}{2}\mathfrak{I} \text{ as we started with}$

$\begin{aligned} \mathfrak{I} = \frac{1}{2}\mathfrak{I} -\pi ln2\implies \mathfrak{I} = \boxed{-\pi ln2}\end{aligned}$

$\large \implies A=1,B=2\implies \boxed{A+B=3}$