sketch the region enclosed by the given curves. y = sec2(x), y = 8 cos(x), −π/3 ≤ x ≤ π/3. The given curves are y=sec2x and y=8cosx with the interval −π3≤x≤π3.The region enclosed by the curves y=sec2x and y=8cosx in the interval −π3≤x≤π3 is sketched as shown below:Sketch of region.Thus, the region is sketched.The area of the region enclosed by y=sec2x and y=8cosx in the interval −π3≤x≤π3 is given by:Area=π3∫−π3 (8cosx−sec2x)dxUse ∫cosxdx=sinx and ∫sec2xdx=tanx and find the area as follows:Area=π3∫−π3 (8cosx−sec2x)dx=π3∫−π3 8cosxdx−π3∫−π3 (sec2x)dx=π3−π3(8sinx)−π3−π3(tanx)Use sin(π3)=√32 , sin(−π3)=−√32 , tan(π3)=√3 and tan(−π3)=−√3 and further simplify the area as follows:Area=π3−π3(8sinx)−π3−π3(tanx)=8(sin(π3)−sin(−π3))−(tan(π3)−tan(−π3))=8(√32−(−√32))−((√3)−(−√3))=8√3−2√3Therefore, Area=6√3.