Evaluate the area between y = sqrt tan x / (cosx)^2, x = 0 , x= pi/4.

To evaluate the area, we'll have to calculate the definite
integral.

Integral (f(x) - ox)dx = Int f(x)dx, where y =
f(x)

We'll apply Leibniz Newton formula to evaluate the
area:

Int f(x)dx = F(b) - F(a), where a and b are the lower
and upper limits. In our case, a = 0 and b = pi/4.

Int
(sqrt tan x)dx / (cosx)^2, from 0 to pi/4 =
F(pi/4)-F(0)

 We notice that if we'll differentiate tan x,
we'll get 1/ (cosx)^2.

So, we'll note tan x =
t

(tan x)'dx = dt

dx/ (cosx)^2
= dt

We'll re-write the
integral:

Int (sqrt tan x)dx / (cosx)^2 = Int (sqrt 
t)dt

Int (sqrt  t)dt = Int [t^(1/2)]dt = t^(1/2 + 1) / (1/2
+ 1)

Int (sqrt  t)dt = t^(3/2) / (3/2) = (2 sqrt
t^3)/3

Int (sqrt  t)dt = (2tsqrt
t)/3

F(pi/4) = [2*(tan pi/4)*sqrt tan pi/4]/3 = 2*1*1/3 =
2/3

F(0) = [2*(tan 0)*sqrt tan 0]/3 = 2*0*0/3 =
0

Int (sqrt tan x)dx / (cosx)^2 = F(pi/4)-F(0) = 2/3 -
0

Int (sqrt tan x)dx / (cosx)^2 =
2/3