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

##### 1 Answer

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!