May be you can have an improper integral

Use $\color{#D61F06}{\int_a^b f(x)dx=\int_a^b f(a+b-x)dx}$ . $\mathfrak{I}=\int_0^{\frac{\pi}{2}}\dfrac{\sin^2 x\, dx}{1+\sin x \cos x } =\int_0^{\pi /2} \dfrac{\cos^2 x \, dx}{1+ \sin x \cos x }$ Adding and using $\color{#D61F06}{2\sin x\cos x=\sin 2x}$ . $\large\implies \mathfrak{I}=\int_0^{\frac{\pi}{2}}\dfrac{\, dx}{2+\sin 2x}$ Using $\color{#D61F06}{\sin 2x=\dfrac{2 \tan x}{1+\tan^2 x}}$ . $\large\implies \mathfrak{I}=\int_0^{\frac{\pi}{2}} \dfrac 12 \left(\dfrac{(1+\tan^2 x)dx}{1+\tan x+\tan^2 x }\right)$ $~~~~~~(\color{#D61F06}{1+\tan^2 x=d(\tan x)})$ $\large =\dfrac 12 \left(\int_0^{\frac{\pi}{2}} \dfrac{d(\tan x)}{(\tan x+\frac 12)^2+\frac 34}\right)$ $\Large =\dfrac{1}{\sqrt 3}\left(\tan^{-1}\left(\dfrac{2\tan x+1}{\sqrt 3}\right)\right)|_{\small 0}^{\small{\frac{\pi}{2}}}$ $\Large =\dfrac{\pi}{3\sqrt 3}$ $\huge \approx \boxed{\color{#007fff}{0.605}}$