Ta có:  fx+f−x=2+2cos2x

=2+22cos2x−1=2cos2x=2cosx

Đặt t = −x Þ dt = − dx Û dx = − dt

Khi đó:

I=∫−3π23π2fxdx=−∫3π2−3π2f−tdt=∫−3π23π2f−tdt=∫−3π23π2f−xdx

⇒2I=∫−3π23π2fxdx+∫−3π23π2f−xdx=∫−3π23π2fx+f−xdx

=∫−3π23π22cosxdx=−2∫−3π23π2cosxdx

=−2sinx−3π23π2=−2sin3π2+2sin−3π2

= (−2).(−1) + 2.1 = 4

Vậy  I=∫−3π23π2fxdx=2.