The value of $$\frac{16}{\pi^3} \int_0^{\frac{\pi}{2}} f(x)g(x) \, dx$$ is ______.

Let us denote$$I=\frac{16}{\pi^{3}}\int_{0}^{\frac{\pi}{2}}f(x)g(x)\,dx=\frac{16}{\pi^{3}}\int_{0}^{\frac{\pi}{2}}\sin^{2}x\,\sqrt{\frac{\pi x}{2}-x^{2}}\;dx\;.$$

Step 1 : Scale the variable
Put $$x=\frac{\pi}{2}\,t,\qquad 0\le t\le 1,\qquad dx=\frac{\pi}{2}\,dt.$$
Then$$\sqrt{\frac{\pi x}{2}-x^{2}}=\sqrt{\frac{\pi}{2}\left(\frac{\pi}{2}t\right)-\left(\frac{\pi}{2}t\right)^{2}}=\frac{\pi}{2}\sqrt{t-t^{2}}=\frac{\pi}{2}\sqrt{t(1-t)}.$$

Substituting these in the integral,

$$\int_{0}^{\frac{\pi}{2}}\sin^{2}x\,\sqrt{\frac{\pi x}{2}-x^{2}}\;dx=\left(\frac{\pi}{2}\right)^{2}\int_{0}^{1}\sin^{2}\!\left(\frac{\pi}{2}t\right)\sqrt{t(1-t)}\;dt$$ $$=\frac{\pi^{2}}{4}\int_{0}^{1}\sin^{2}\!\left(\frac{\pi}{2}t\right)\sqrt{t(1-t)}\;dt\;.$$ Therefore

$$I=\frac{16}{\pi^{3}}\cdot\frac{\pi^{2}}{4}\int_{0}^{1}\sin^{2}\!\left(\frac{\pi}{2}t\right)\sqrt{t(1-t)}\;dt=\frac{4}{\pi}\int_{0}^{1}\sin^{2}\!\left(\frac{\pi}{2}t\right)\sqrt{t(1-t)}\;dt\;.$$

Step 2 : Replace $$\sin^{2}$$ by a cosine
The identity $$\sin^{2}y=\frac{1-\cos2y}{2}$$ gives $$\sin^{2}\!\left(\frac{\pi}{2}t\right)=\frac{1-\cos(\pi t)}{2}.$$

Hence

$$I=\frac{4}{\pi}\int_{0}^{1}\frac{1-\cos(\pi t)}{2}\;\sqrt{t(1-t)}\;dt =\frac{2}{\pi}\Bigl[I_{1}-I_{2}\Bigr],$$ where

$$I_{1}=\int_{0}^{1}\sqrt{t(1-t)}\;dt,\qquad I_{2}=\int_{0}^{1}\cos(\pi t)\sqrt{t(1-t)}\;dt.$$

Step 3 : Evaluate $$I_{1}$$ using the Beta function
Write $$\sqrt{t(1-t)}=t^{\frac12}(1-t)^{\frac12}.$$
Hence $$I_{1}=B\!\left(\frac32,\frac32\right)=\frac{\Gamma\!\left(\frac32\right)\Gamma\!\left(\frac32\right)}{\Gamma(3)}.$$ Using $$\Gamma\!\left(\frac12\right)=\sqrt{\pi},\quad \Gamma\!\left(\frac32\right)=\frac12\sqrt{\pi},\quad \Gamma(3)=2,$$ we get $$I_{1}=\frac{\left(\dfrac12\sqrt{\pi}\right)^{2}}{2} =\frac{\pi/4}{2}=\frac{\pi}{8}.$$

Step 4 : Show that $$I_{2}=0$$ by symmetry
Consider the substitution $$t=1-u.$$ Then $$dt=-du$$ and $$\sqrt{t(1-t)}=\sqrt{(1-u)u}=\sqrt{u(1-u)}.$$ Thus $$I_{2}=\int_{0}^{1}\cos(\pi t)\sqrt{t(1-t)}\;dt =\int_{1}^{0}\cos\!\bigl(\pi(1-u)\bigr)\sqrt{u(1-u)}(-du)$$ $$=\int_{0}^{1}\cos\!\bigl(\pi-\pi u\bigr)\sqrt{u(1-u)}\;du =\int_{0}^{1}\bigl[-\cos(\pi u)\bigr]\sqrt{u(1-u)}\;du=-I_{2}.$$
Hence $$I_{2}=-I_{2}\;\Longrightarrow\;I_{2}=0.$$

Step 5 : Assemble the result
$$I=\frac{2}{\pi}\Bigl[I_{1}-I_{2}\Bigr]=\frac{2}{\pi}\cdot\frac{\pi}{8}= \frac14 = 0.25.$$

Therefore, the required value is 0.25.