Question

What is the area between `y = secx` and `y = cosx` on `[-pi/4, pi/4]`?

Answer 1

As you can see in the following graph, the cosine graph lies above the secant graph in the given interval.

So our integral for area will be

`A = int_(-pi/4)^(pi/4) 2cosx - secxdx`

These are two known integrals.

`A = [2sinx - ln|tanx + secx|]_(-pi/4)^(pi/4)`

`A = 2sin(pi/4) - ln|tan(pi/4) + 1/cos(pi/4)| - 2sin(-pi/4) + ln|tan(-pi/4) + sec(-pi/4)|`

`A = 2/sqrt(2) -ln|1 + sqrt(2)| + 2/sqrt(2) + ln|-1 + sqrt(2)|`

`A = 4/sqrt(2) + ln((-1 + sqrt(2))/(1 + sqrt(2)))`

Hopefully this helps!