Para calcular a integral \( \int_{0}^{\pi/2} \cos^2(x) \sin^2(x) \, dx \), podemos usar a identidade trigonométrica: \[ \sin^2(x) = \frac{1 - \cos(2x)}{2} \quad \text{e} \quad \cos^2(x) = \frac{1 + \cos(2x)}{2}. \] Assim, temos: \[ \cos^2(x) \sin^2(x) = \left(\frac{1 + \cos(2x)}{2}\right) \left(\frac{1 - \cos(2x)}{2}\right) = \frac{(1 + \cos(2x))(1 - \cos(2x))}{4} = \frac{1 - \cos^2(2x)}{4} = \frac{\sin^2(2x)}{4}. \] Portanto, a integral se torna: \[ \int_{0}^{\pi/2} \cos^2(x) \sin^2(x) \, dx = \int_{0}^{\pi/2} \frac{\sin^2(2x)}{4} \, dx = \frac{1}{4} \int_{0}^{\pi/2} \sin^2(2x) \, dx. \] Usando a identidade \( \sin^2(a) = \frac{1 - \cos(2a)}{2} \): \[ \int_{0}^{\pi/2} \sin^2(2x) \, dx = \int_{0}^{\pi/2} \frac{1 - \cos(4x)}{2} \, dx = \frac{1}{2} \left[ x - \frac{\sin(4x)}{4} \right]_{0}^{\pi/2} = \frac{1}{2} \left[ \frac{\pi}{2} - 0 \right] = \frac{\pi}{4}. \] Assim, substituindo de volta: \[ \int_{0}^{\pi/2} \cos^2(x) \sin^2(x) \, dx = \frac{1}{4} \cdot \frac{\pi}{4} = \frac{\pi}{16}. \] Portanto, a resposta é: \[ \int_{0}^{\pi/2} \cos^2(x) \sin^2(x) \, dx = \frac{\pi}{16}. \]