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

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.

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