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

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.