Cho y=f(x+pi/2) là hàm chẵn trên -pi/2 và pi/2

* Hướng dẫn giải

Từ $f\left(x\right)+f\left(x+\frac{\mathrm{\pi }}{2}\right)=\sin \left(x\right)+\cos \left(x\right)$ cho $x=\frac{\mathrm{\pi }}{2},x=\frac{-\mathrm{\pi }}{2}$ ta có

$\left\{\begin{array}{l}f\left(\frac{\mathrm{\pi }}{2}\right)+f\left(\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{2}\right)=\sin \left(\frac{\mathrm{\pi }}{2}\right)+\cos \left(\frac{\mathrm{\pi }}{2}\right)=\sin \left(\frac{\mathrm{\pi }}{2}\right)\\ f\left(\frac{-\mathrm{\pi }}{2}\right)+f\left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{2}\right)=\sin \left(\frac{-\mathrm{\pi }}{2}\right)+\cos \left(\frac{\mathrm{\pi }}{2}\right)=\sin \left(\frac{\mathrm{\pi }}{2}\right)\end{array}\right.$

Chú ý do $y=f\left(x+\frac{\mathrm{\pi }}{2}\right)$ là hàm chẵn trên $\left[\frac{-\mathrm{\pi }}{2};\frac{\mathrm{\pi }}{2}\right]$ nên $f\left(\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{2}\right)$ = $f\left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{2}\right)$

$$\begin{gathered} \Rightarrow f\left(\frac{\mathrm{\pi }}{2}\right)-f\left(\frac{-\mathrm{\pi }}{2}\right)=\sin \left(\frac{\mathrm{\pi }}{2}\right)-\sin \left(-\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow f\left(x\right)=\sin \left(x\right) \end{gathered}$$

Vậy ${\int }_0^{\frac{\mathrm{\pi }}{2}}f\left(x\right)dx$ = ${\int }_0^{\frac{\mathrm{\pi }}{2}}\sin \left(x\right)=1$

Đáp án cần chọn là B