Question:

Let the given expression be $I$: $$I = \int_{0}^{\pi/2} (2\sin^2 x - 1) g(x) dx = -\int_{0}^{\pi/2} \cos(2x) \sqrt{\dfrac{\pi x}{2} - x^2} dx$$ Let's apply the substitution $u = 2x - \dfrac{\pi}{2} \implies dx = \dfrac{du}{2}$: - When $x = 0$, $u = -\dfrac{\pi}{2}$. - When $x = \dfrac{\pi}{2}$, $u = \dfrac{\pi}{2}$. Now rewrite the term inside the square root: $$\dfrac{\pi x}{2} - x^2 = x\left(\dfrac{\pi}{2} - x\right) = \left(\dfrac{\pi}{4} + \dfrac{u}{2}\right)\left(\dfrac{\pi}{4} - \dfrac{u}{2}\right) = \dfrac{\pi^2}{16} - \dfrac{u^2}{4}$$ which is an even function of $u$. Rewrite the cosine term: $$-\cos(2x) = -\cos\left(u + \dfrac{\pi}{2}\right) = \sin u$$ which is an odd function of $u$. Thus, the integrand is: $$\sin u \sqrt{\dfrac{\pi^2}{16} - \dfrac{u^2}{4}}$$ This is the product of an odd function and an even function, which is an odd function. Since the integration limits are symmetric $[-\pi/2, \pi/2]$, the integral of this odd function is 0. Thus, the value of the expression is 0.