Question

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

Hint:

For complex integrals, consider symmetry and known integral formulas like \( \int \sqrt{a^2 - x^2} \, dx \).

Answer 1

Given \( f(x) = \sin^2(x) \), \( g(x) = \sqrt{\frac{\pi x}{2} - x^2} \), and: \[ I = \frac{16}{\pi^3} \int_0^{\pi/2} \sin^2(x) \sqrt{\frac{\pi x}{2} - x^2} \, dx. \] Using the substitution \( \sin^2(x) + \cos^2(x) = 1 \): \[ I = \frac{16}{\pi^3} \int_0^{\pi/2} \cos^2(x) \sqrt{\frac{\pi x}{2} - x^2} \, dx. \] Combining: \[ 2I = \frac{16}{\pi^3} \int_0^{\pi/2} \left(\sin^2(x) + \cos^2(x)\right) \sqrt{\frac{\pi x}{2} - x^2} \, dx. \] Simplifying: \[ 2I = \frac{16}{\pi^3} \cdot \frac{\pi^3}{32} = \frac{1}{4}. \] Thus: \[ I = 0.25. \]