what is the area between the curves 3cos(x) and 3sec^2(x) between -pi/4,pi/4

Question

area between curves, 3cos(x) and 3sec^2(x) between -pi/4 and pi/4

Answer 1

Because of symmetry we only need to consider the enclosed area between x=0 and x=π/4, and the curves touch when cos(x)=sec²(x), cos³(x)=1, cos(x)=1, x=0.

Area=6∫(sec²(x)-cos(x))dx[0,π/4]=

6[tan(x)-sin(x)][0,π/4]=6(1-√2/2)=6-3√2=1.757 approx.