MathematicsEasy210×since 2002Q3466Let f(x)f(x)f(x) be a function satisfying f(x)+f(π−x)=π2,∀x∈Rf(x)+f(\pi-x)=\pi^{2}, \forall x \in \mathbb{R}f(x)+f(π−x)=π2,∀x∈R. Then \int_\limits{0}^{\pi} f(x) \sin x d x is equal to :Aπ2\pi^{2}π2Bπ22\frac{\pi^{2}}{2}2π2C2π22 \pi^{2}2π2Dπ24\frac{\pi^{2}}{4}4π2Check answerSkip