Questão: Qual é o valor da integral definida de cos^2(x)dx no intervalo de 0 a π/2? a) π/2 b) π/4 c) 1 d) 0

Para calcular a integral definida de \(\cos^2(x)\) no intervalo de \(0\) a \(\frac{\pi}{2}\), podemos usar a identidade trigonométrica: \[ \cos^2(x) = \frac{1 + \cos(2x)}{2} \] Assim, a integral se torna: \[ \int_0^{\frac{\pi}{2}} \cos^2(x) \, dx = \int_0^{\frac{\pi}{2}} \frac{1 + \cos(2x)}{2} \, dx \] Podemos separar a integral: \[ = \frac{1}{2} \int_0^{\frac{\pi}{2}} 1 \, dx + \frac{1}{2} \int_0^{\frac{\pi}{2}} \cos(2x) \, dx \] Calculando a primeira parte: \[ \frac{1}{2} \int_0^{\frac{\pi}{2}} 1 \, dx = \frac{1}{2} \left[ x \right]_0^{\frac{\pi}{2}} = \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4} \] Agora, a segunda parte: \[ \frac{1}{2} \int_0^{\frac{\pi}{2}} \cos(2x) \, dx = \frac{1}{2} \left[ \frac{\sin(2x)}{2} \right]_0^{\frac{\pi}{2}} = \frac{1}{2} \left( \frac{\sin(\pi) - \sin(0)}{2} \right) = \frac{1}{2} \cdot 0 = 0 \] Portanto, somando as duas partes: \[ \int_0^{\frac{\pi}{2}} \cos^2(x) \, dx = \frac{\pi}{4} + 0 = \frac{\pi}{4} \] Assim, a resposta correta é: b) \(\frac{\pi}{4}\).