A basic integral

$\because \int_{a}^b f(x) \, dx=\int_{a}^b f(a+b-x) \, dx$ $\therefore \Large \mathfrak{I}=\displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{2} x \, dx=\displaystyle \int_{0}^{\frac{\pi}{2}} \cos^{2} x \, dx$ $~~~~~~~~~~~~~~~(\color{#D61F06}{\sin (\frac{\pi}{2}-x)=\cos x})~~~~\\$
Adding we get: $\Large 2\mathfrak{I}=\int_{0}^{\frac{\pi}{2}} (1) \, dx=\frac{\pi}{4}$ $\huge \mathfrak{I}=\boxed{\color{#007fff}{\frac{\pi}{4}}}$

Alternate method: Use $\sin^2 x=\frac{1-\cos 2x}{2}$ so that integral simplifies to: $\Large \int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2x}{2}$ $\Large=\frac{1}{4}(2x-\sin 2x)\huge{|}_{\small 0}^{\small \frac{\pi}{2} }$ $\huge=\frac{\pi}{4}$

$\frac{\pi}{2} = \int_{0}^{\frac{\pi}{2}} {1} dx = \int_{0}^{\frac{\pi}{2}} dx =$ $\int_{0}^{\frac{\pi}{2}} {cos^2(x)} dx + \int_{0}^{\frac{\pi}{2}} {sin^2(x)} dx$ ;Now

$\int_{0}^{\frac{\pi}{2}} {sin^2(x)} dx =$ Integrating by parts $= \left[-sin(x)cos(x) \right]_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} {cos^2(x)} dx =$ $=\int_{0}^{\frac{\pi}{2}} {cos^2(x)} dx \Rightarrow \frac{\pi}{2} = 2 \int_{0}^{\frac{\pi}{2}} {sin^2(x)} dx \Rightarrow$ $\frac{\pi}{4} = \int_{0}^{\frac{\pi}{2}} {sin^2(x)} dx$