Evaluate Definite Integral Difference of Product and Single Function

To find the value of $I = 2 \int_{0}^{\frac{\pi}{2}} f(x) g(x) \, dx - \int_{0}^{\frac{\pi}{2}} g(x) \, dx$, we simplify the expression using the definition $f(x) = \sin^2 x$:

$I = \int_{0}^{\frac{\pi}{2}} \left(2 \sin^2 x - 1\right) g(x) \, dx = -\int_{0}^{\frac{\pi}{2}} \cos(2x) g(x) \, dx$

where $g(x) = \sqrt{\frac{\pi x}{2} - x^2}$.

We evaluate the integral $I_1 = \int_{0}^{\frac{\pi}{2}} \cos(2x) g(x) \, dx$ using the property $\int_{a}^{b} h(x) \, dx = \int_{a}^{b} h(a+b-x) \, dx$.

Replacing $x$ with $\frac{\pi}{2} - x$, we get:

$g\left(\frac{\pi}{2} - x\right) = \sqrt{\frac{\pi}{2}\left(\frac{\pi}{2} - x\right) - \left(\frac{\pi}{2} - x\right)^2} = \sqrt{\frac{\pi x}{2} - x^2} = g(x)$

$\cos\left(2\left(\frac{\pi}{2} - x\right)\right) = \cos(\pi - 2x) = -\cos(2x)$

Thus, the integral becomes:

$I_1 = \int_{0}^{\frac{\pi}{2}} -\cos(2x) g(x) \, dx = -I_1$

$2 I_1 = 0 \implies I_1 = 0$

Therefore, $I = -I_1 = 0$.