Para calcular a integral \( \int_0^{\frac{\pi}{2}} \sin(x) \cos(x) \, dx \), podemos usar a identidade trigonométrica: \[ \sin(x) \cos(x) = \frac{1}{2} \sin(2x) \] Assim, a integral se torna: \[ \int_0^{\frac{\pi}{2}} \sin(x) \cos(x) \, dx = \int_0^{\frac{\pi}{2}} \frac{1}{2} \sin(2x) \, dx \] Agora, podemos fatorar a constante \( \frac{1}{2} \) para fora da integral: \[ = \frac{1}{2} \int_0^{\frac{\pi}{2}} \sin(2x) \, dx \] A integral de \( \sin(2x) \) é: \[ -\frac{1}{2} \cos(2x) \] Portanto, avaliando a integral: \[ = \frac{1}{2} \left[-\frac{1}{2} \cos(2x)\right]_0^{\frac{\pi}{2}} = \frac{1}{2} \left[-\frac{1}{2} (\cos(\pi) - \cos(0))\right] \] Sabendo que \( \cos(\pi) = -1 \) e \( \cos(0) = 1 \): \[ = \frac{1}{2} \left[-\frac{1}{2} (-1 - 1)\right] = \frac{1}{2} \left[-\frac{1}{2} (-2)\right] = \frac{1}{2} \cdot 1 = \frac{1}{2} \] Portanto, a resposta correta é: a) \( \frac{1}{2} \)