Chọn B
\(I=\int _{0}^{\frac{\pi }{2} }\sin ^{2} x{\rm d}x =\int _{0}^{\frac{\pi }{2} }\frac{1-\cos 2x}{2} {\rm d}x=\left. \frac{1}{2} \left(x-\frac{\sin 2x}{2} \right)\right|_{0}^{\frac{\pi }{2} } =\frac{\pi }{4} . \)
\(J=\int _{0}^{\frac{\pi }{2} }\cos ^{2} x{\rm d}x =\int _{0}^{\frac{\pi }{2} }\frac{1+\cos 2x}{2} {\rm d}x=\left. \frac{1}{2} \left(x+\frac{\sin 2x}{2} \right)\right|_{0}^{\frac{\pi }{2} } =\frac{\pi }{4} . \)
Vậy I=J.