Ta có: f (x) + f (−x) = 3 − 2cos x

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

=∫−π2π23−2cosxdx=3x−2sinx−π2π2

=3 . π2−2sinπ2−3 . −π2+2 sin−π2

= 3p − 4

Vậy   I=∫−π2π2fxdx=3π2−2.