y = 5cos x, y = (3sec(x))^2, x = -pi/4, x = pi/4?

Question

Find the area of the region.

(Sketh the region enclosed by the given curves. Decide whether to integrate with respect to x or y).

Answer 1

The second curve is (3 sec x)^2 = 9 sec^2 x. In the given region this is always at least 9 and hence always greater than the other curve which is at most 5. So we have
∫(-π/4 to π/4) (9 sec^2 x - 5 cos x) dx
= [9 tan x - 5 sin x] [-π/4 to π/4]
= (9 - 5/√2) - (-9 + 5/√2)
= 18 - 10√2.