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

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### 3 Answers

y= sqrt(tanx)/(cosx)^2 x = 0 and x= pi/4

We know that the area under y is intergral y = F(x)

integrall y = intg (sqrt(tanx))/cos^2

= (2/3) tanx *sqrt(tanx)

Now the area is:

A = F(pi/4) - F(0)

= 2*tan(pi/4)*sqrt(tan(pi/4)) - 2*tan0 * sqrt(tan0)

= (2/3)*1* 1 - 2*0*0 = 2/3

Then the area A = 2/3

To find Integral {sqrt tanx/(cosx)^2} dx x = 0 to x= pi/4.

Let us have the transformation: tanx = t.

Then (tanx)' dx = (secx)^2 dx = dt. Or

dx/cosx)^2 = dt. When x = 0, tanx = t = 0. When x = pi/4, t = tanpi/4 = 1.

So the given integral becomes:

Int sqrt tanx (dx/cosx)^2 = Int ( sqrt t ) dt = {t^(1/2+1)}/(1/2+1)

Taking the limits at t = 1 and 0, for {t^(3/2)}(2/3) we get:

(1^(3/2) (2/3) - 0

2/3.

So the Int {sqrt tanx/ cosx)^2 dx from x= 0 to pi/4 } = 2/3.

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**