Find the area. y=2sec(pi*x/6), x=0, x=2, y=0?

Question

need all steps please.

Answer 1

The area is just the definite integral:
A = [integral sign] [0 to 2] 2sec(pi*x/6)dx
The question is how the integral can be evaluated.
First let us change the variable: z = pi*x/12, and so dx = (12/pi)dz
A = (12/pi)*[integral sign] [0 to pi/6] 2sec 2z dz
Second let us notice:
2sec 2z = 2/ cos 2z = 2/(cos^2 z - sin^2 z)
= (2/cos^2 z)/ (1 - tan^2 z)
=sec^2 z{ 1/(1 + tan z) + 1/(1 - tan z) }
and also realize that d(tan z)/dz = sec^2 z
Hence the integral can be easily evaluated as:
A = (12/pi)*{ Ln[(1 + tan z) /(1 - tan z) ] } |0 to pi/6
= (12/pi)*{ Ln[(1 + tan(pi/6)) /(1 - tan(pi/6)) ] }
= (12/pi)*{ Ln[(1 + sqrt(3)/3) /(1 - sqrt(3)/3) ] }
= (12/pi)*{ Ln[(3 + sqrt(3)) /(3 - sqrt(3)) ] }
= (12/pi)*{ Ln(2 + sqrt(3)) }