What is the area of the region between the function sin^2x/cos^6x, x axis and x=0 to x=pi/4?

To determine the area of the region bounded by the given
curve, x axis and the lines x = 0 and x = pi/4, we'll have to calculate the definite
integral of the given function, using Leibniz-Newton
formula.

To calculate the definite integral, first, we'll
change the variable using the substitution tan x = t.

We
also use the identity:

1 + (tan x)^2 = 1/(cos
x)^2

The given function
is:

(sin x)^2/(cos x)^6 = [(sin x)^2/(cos x)^4]*[1/(cos
x)^2]

(sin x)^2/(cos x)^6 = (tan x)^2*(tan x)'/(cos
x)^2

(sin x)^2/(cos x)^6 = (tan x)^2*[1 + (tan x)^2]*(tan
x)'

We'll integrate both
sides:

Int (sin x)^2dx/(cos x)^6 = Int (tan x)^2*[1 + (tan
x)^2]*(tan x)'dx

Int (sin x)^2dx/(cos x)^6 = Int (t^2 +
t^4)dt

Int (t^2 + t^4)dt  =Int t^2dt + Int
t^4dt

Int (t^2 + t^4)dt  = t^3/3 +
t^5/5

Int (sin x)^2dx/(cos x)^6 = (tan x)^3/3 + (tan
x)^5/5

We'll apply Leibniz-Newton
formula:

Int (sin x)^2dx/(cos x)^6 = F(pi/4)-
F(0)

F(pi/4) = (tan pi/4)^3/3 + (tan
pi/4)^5/5

F(pi/4) = 1/3 +
1/5

F(0) = 0

Int (sin
x)^2dx/(cos x)^6 = 1/3 + 1/5

Int (sin x)^2dx/(cos x)^6 =
8/15

The area of the region bounded by the
given curve, x axis and the lines x = 0 and x = pi/4 is Int (sin x)^2dx/(cos x)^6 = 8/15
square units.