Problema 79: Calcule \integral_0^{\frac{\pi}{2}} \sin^2(x) \, dx. a) \frac{\pi}{4} b) \frac{1}{2} c) \frac{1}{3} d) 0

Para calcular a integral \(\int_0^{\frac{\pi}{2}} \sin^2(x) \, dx\), podemos usar a identidade trigonométrica: \[ \sin^2(x) = \frac{1 - \cos(2x)}{2} \] Substituindo na integral, temos: \[ \int_0^{\frac{\pi}{2}} \sin^2(x) \, dx = \int_0^{\frac{\pi}{2}} \frac{1 - \cos(2x)}{2} \, dx \] Isso se separa em duas integrais: \[ = \frac{1}{2} \int_0^{\frac{\pi}{2}} 1 \, dx - \frac{1}{2} \int_0^{\frac{\pi}{2}} \cos(2x) \, dx \] Calculando a primeira integral: \[ \int_0^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_0^{\frac{\pi}{2}} = \frac{\pi}{2} \] Agora, a segunda integral: \[ \int_0^{\frac{\pi}{2}} \cos(2x) \, dx = \left[ \frac{\sin(2x)}{2} \right]_0^{\frac{\pi}{2}} = \frac{\sin(\pi)}{2} - \frac{\sin(0)}{2} = 0 \] Portanto, a integral se torna: \[ = \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4} \] Assim, a resposta correta é: a) \(\frac{\pi}{4}\)