# What is the area of the region closed by the curve y=1/cos^2x, x axis and the lines x=pi/4 , x=pi/3?

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We need to find the area of the region enclosed between the curve y = 1/(cos)^2 , the x-axis and the lines x = pi/4 and x = pi/3

To do this the definite integral of y = 1/(cos x)^2 has to be found between the limits x = pi/4 and x = pi/3.

1/(cos x)^2 = (sec x)^2

Int [ (sec x)^2 dx], x = pi/4 to x = pi/3

=> tan (pi/3) - tan (pi/4)

=> sqrt 3 - 1

**The required are is sqrt 3 - 1**

To determine the area of he region bounded by the given curve and lines, we'll have to compute the definite integral of y.

This integral will be evaluated using the Leibniz-Newton formula.

Int f(x)dx = F(b) - F(a), where x = a to x = b

Let y = f(x) = 1/(cos x)^2

We'll compute the indefinite integral, first:

Int dx/(cos x)^2 = tan x + C

We'll note the result F(x) = tan x + C

We'll determine F(a), for a = pi/3:

F(pi/3) = tan pi/3

F(pi/3) = sqrt 3

We'll determine F(b), for b = pi/4:

F(pi/4) = tan pi/4

F(pi/4) = 1

We'll evaluate the definite integral:

Int dx/(cos x)^2 = F(pi/3) - F(pi/4)

Int dx/(cos x)^2 = sqrt 3 - 1

**The area of the region, bounded by the curve y=1/(cos x)^2, x axis and the lines x=pi/4 and x=pi/3, is: A = (sqrt 3 - 1) square units.**