Evaluate the area of the surface under the curve . Y=1/(sin^2x+4cos^2x+2), x=0 and x=pi/4.

giorgiana1976 | Student

Before evaluating the definite integral, to determine the area located under the curve, the given lines and x axis, we verify if the function is an even function.

A function is even if f(-x) = f(x)

We'll substitute x by -x:

f(-x) = 1/{[sin(-x)]^2+4[cos(-x)]^2+2}

f(-x) = 1/{[sin(x)]^2+4[cos(x)]^2+2} = f(x)

Since the function is even, we'll suggest the substitution tan x = t.

First, we'll factorize the denominator by [cos(x)]^2:

 1/[cos(x)]^2{[tan(x)]^2 +4 + 2/[cos(x)]^2} =  1/[cos(x)]^2{[tan(x)]^2 +4 + 2*[1+(tanx)^2]}

Since tan x = t, then  dx/(cos x)^2 = dt

We'll re-write the integral in t:

(1/3)Int dt/(t^2 + 2) = (1/3*sqrt2)*arctan (t/sqrt2)

Int f(x)dx = (1/3*sqrt2)*arctan [(tan pi/4)/sqrt2] - (1/3*sqrt2)*arctan [(tan 0)/sqrt2]

Int f(x)dx = (sqrt2/6)*arctan (sqrt2/2)

The area of the surface is A = (sqrt2/6)*arctan (sqrt2/2) =  0.1849 square units approx.

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