Ta có: f (x) + f (−x) = 2cos 2x

Đặt t = −x Þ dt = − dx Û dx = − dt

Khi đó:

I=∫−π2π2fxdx=−∫π2−π2f−tdt=∫−π2π2f−tdt=∫−π2π2f−xdx

⇒2I=∫−π2π2fxdx+∫−π2π2f−xdx=∫−π2π2fx+f−xdx

=∫−π2π22cos2xdx=2∫−π2π2cos2xdx

=sin2x−π2π2=sinπ−sin−π = 0 

Vậy  I=∫−π2π2fxdx=0.