Question

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

Hint:

Use symmetry and trigonometric identities to simplify integrals involving squares of trigonometric functions.

Answer 1

Let: \[ I = \int_0^{\pi/2} f(x)g(x) \, dx. \] Substituting \( f(x) = \sin^2(x) \) and \( g(x) = \sqrt{\frac{\pi x}{2} - x^2} \): \[ I = \int_0^{\pi/2} \sin^2(x) \sqrt{\frac{\pi x}{2} - x^2} \, dx. \] Using the trigonometric identity \( \sin^2(x) = 1 - \cos^2(x) \): \[ I = \int_0^{\pi/2} \cos^2(x) \sqrt{\frac{\pi x}{2} - x^2} \, dx. \] Combining integrals: \[ 2I = \int_0^{\pi/2} \left(\sin^2(x) + \cos^2(x)\right) \sqrt{\frac{\pi x}{2} - x^2} \, dx = \int_0^{\pi/2} g(x) \, dx. \] Thus: \[ 2 \int_0^{\pi/2} f(x)g(x) \, dx - \int_0^{\pi/2} g(x) \, dx = 0. \]