Find value of $$I=\int_{0}^{\frac{\pi}{2}} \log ^2(\sin x)dx \tag{1}$$
I tried using property $\int_{a}^{b} f(x)dx=\int_{a}^{b}f(a+b-x)dx$ we get
$$I=\int_{0}^{\frac{\pi}{2}}\log^2(\cos x)dx \tag{2}$$
Adding $(1)$ and $(2)$ we get
$$2I=\int_{0}^{\frac{\pi}{2}}\left(\log(\sin x)+\log(\cos x)\right)^2-2 \log(\sin x)\log(\cos x)dx$$
any clue here?
