Let \(f:\left[0, \frac{\pi}{2}\right] \rightarrow[0,1]\) be the function defined by \(f(x)=\sin ^2 x\) and let \(g:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty)\) be the function defined by \(g(x)=\sqrt{\frac{\pi x}{2}-x^2}\).
The value of \(\frac{16}{\pi^3} \int_0^{\frac{\pi}{2}} f(x) g(x) d x\) is

Now \(I_1=\int_0^{\frac{\pi}{2}} f(x) \cdot g(x) d x=\frac{1}{2} \int_0^{\frac{\pi}{2}} g(x) d x\) (it is explained in previous question solution)
i.e. \(\frac{1}{2} \int_0^{\frac{\pi}{2}} \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2} \mathrm{dx}\)
Using \(\int \sqrt{a^2-x^2}=\frac{1}{2}\left(x \sqrt{a^2-x^2}+a^2 \sin ^{-1}\left(\frac{x}{a}\right)\right)+C\)
\(\Rightarrow \frac{1}{2}\left[\frac{\left(x-\frac{\pi}{4}\right)}{2} \sqrt{\frac{\pi x}{2}-x^2}+\frac{\frac{\pi^2}{2}}{2} \sin ^{-1}\left(\frac{x-\frac{\pi}{4}}{\frac{\pi}{4}}\right)\right]_0^{\pi / 2}\)
\(\begin{aligned} & \Rightarrow \frac{1}{2}\left[\left(0+\frac{\pi^3}{64}\right)-\left(0+\left(\frac{-\pi^3}{64}\right)\right)\right] \\ & \Rightarrow \frac{1}{2} \times \frac{\pi^3}{32}\end{aligned}\)
Now \(\frac{16}{\pi^3} \times \frac{\pi^3}{64}=\frac{1}{4}=0.25\)