By expressing cos(2x) in terms of cos(x) find the exact value of the integral of cos(2x)/cos^2(x) between the bounds pi/4 and pi/3.
cos(2x)=cos2(x)-sin2(x)=2cos2(x)-1
Therefore:cos(2x)/cos2(x)=(2cos2(x)-1)/cos2(x)=2cos2(x)/cos2(x) - 1/cos2(x)=2 - 1/cos2(x)=2 - sec2(x)
Integral of sec2(x) = tan(x)
Integral of 2 = 2x
[2x - tan(x)] between pi/4 and pi/3
= (2pi/3 - tan(pi/3)) -(pi/2 - tan(pi/4))
= (2pi/3 - sqrt(3)) - (pi/2 - 1)
= pi/6 - sqrt(3) + 1
HF
