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

We have to show that
$$2\int_{0}^{\frac{\pi}{2}} f(x)\,g(x)\,dx-\int_{0}^{\frac{\pi}{2}} g(x)\,dx=0,$$
where $$f(x)=\sin^{2}x$$ and $$g(x)=\sqrt{\dfrac{\pi x}{2}-x^{2}}=\sqrt{x\!\left(\dfrac{\pi}{2}-x\right)}.$$

Step 1: Evaluate $$\displaystyle\int_{0}^{\frac{\pi}{2}} g(x)\,dx$$

Put $$x=\dfrac{\pi}{2}\,t,\;0\le t\le1.$$
Then $$dx=\dfrac{\pi}{2}\,dt$$ and
$$g(x)=\sqrt{\dfrac{\pi x}{2}-x^{2}}=\sqrt{\dfrac{\pi^{2}}{4}\,t-\dfrac{\pi^{2}}{4}\,t^{2}}=\dfrac{\pi}{2}\sqrt{t(1-t)}.$$

Thus
$$\int_{0}^{\frac{\pi}{2}} g(x)\,dx =\int_{0}^{1}\dfrac{\pi}{2}\sqrt{t(1-t)}\;\dfrac{\pi}{2}\,dt =\dfrac{\pi^{2}}{4}\int_{0}^{1}t^{\frac12}(1-t)^{\frac12}\,dt.$$

The integral is the Beta-function $$B\!\left(\tfrac32,\tfrac32\right)=\dfrac{\Gamma\!\left(\tfrac32\right)^2}{\Gamma(3)}.$$
Using $$\Gamma\!\left(\tfrac32\right)=\dfrac{\sqrt{\pi}}{2}$$ and $$\Gamma(3)=2,$$
$$B\!\left(\tfrac32,\tfrac32\right)=\dfrac{\pi/4}{2}=\dfrac{\pi}{8}.$$

Hence
$$\int_{0}^{\frac{\pi}{2}} g(x)\,dx =\dfrac{\pi^{2}}{4}\cdot\dfrac{\pi}{8} =\dfrac{\pi^{3}}{32}.\quad -(1)$$

Step 2: Evaluate $$\displaystyle\int_{0}^{\frac{\pi}{2}}\!f(x)\,g(x)\,dx$$

With the same substitution $$x=\dfrac{\pi}{2}t,$$

$$f(x)\,g(x)=\sin^{2}\!\left(\dfrac{\pi t}{2}\right)\! \left(\dfrac{\pi}{2}\sqrt{t(1-t)}\right),\qquad dx=\dfrac{\pi}{2}\,dt.$$

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

Step 3: Simplify the inner integral

Write $$\sin^{2}\theta=\dfrac{1-\cos2\theta}{2}.$$
Thus

$$\sin^{2}\!\left(\dfrac{\pi t}{2}\right) =\dfrac{1-\cos(\pi t)}{2}.$$

Let
$$I=\int_{0}^{1}\sin^{2}\!\left(\dfrac{\pi t}{2}\right)\sqrt{t(1-t)}\,dt =\dfrac12\int_{0}^{1}\sqrt{t(1-t)}\,dt -\dfrac12\underbrace{\int_{0}^{1}\cos(\pi t)\sqrt{t(1-t)}\,dt}_{K}.$$

Step 4: Show $$K=0$$ by symmetry

Set $$t=\dfrac12+u,\;dt=du,$$ so $$u\in\left[-\dfrac12,\dfrac12\right].$$
Then $$\sqrt{t(1-t)}=\sqrt{\dfrac14-u^{2}},$$ an even function of $$u$$, while

$$\cos(\pi t)=\cos\!\left(\dfrac{\pi}{2}+\pi u\right)=-\sin(\pi u),$$ which is an odd function of $$u$$.

The product $$-\sin(\pi u)\sqrt{\dfrac14-u^{2}}$$ is odd, hence its integral over the symmetric interval vanishes: $$K=0.$$

Step 5: Evaluate $$I$$ and then the required integral

Using $$K=0$$ and $$\displaystyle\int_{0}^{1}\sqrt{t(1-t)}\,dt=\dfrac{\pi}{8},$$

$$I=\dfrac12\cdot\dfrac{\pi}{8}=\dfrac{\pi}{16}.$$

Substituting into $$(2):$$

$$\int_{0}^{\frac{\pi}{2}}\!f(x)g(x)\,dx =\dfrac{\pi^{2}}{4}\cdot\dfrac{\pi}{16} =\dfrac{\pi^{3}}{64}.\quad -(3)$$

Step 6: Form the required expression

$$2\int_{0}^{\frac{\pi}{2}} f(x)g(x)\,dx -\int_{0}^{\frac{\pi}{2}} g(x)\,dx =2\left(\dfrac{\pi^{3}}{64}\right)-\dfrac{\pi^{3}}{32} =\dfrac{\pi^{3}}{32}-\dfrac{\pi^{3}}{32}=0.$$

Hence the required value is 0.