Calculate area bounded by the curve y=cosx/(sin^2x-4), x axis and the lines x = 0 and x = pi/2 .

To calculate the area located under the given curve and
between the given lines, we'll have to calculate the definite integral of the function y
= f(x), whose limits of integration are x = 0 and x =
pi/2.

Int f(x)dx = Int cos x dx/[(sin x)^2 -
4]

We'll use substitution technique and w'ell replace sin x
by another variable, t.

sin x =
t

We'll differentiate both
sides:

cos x dx = dt

We'll
re-write the integral of the function in t:

Int dt/(t^2 -
4) = F(1) - F(0) (Leibniz Newton)

We've replaced the limits
of integration, too.

For x = 0 => sin 0 = t1 =
0

For x = pi/2 => sin pi/2 = t2 =
1

We'll calculate Int dt/(t^2 -
4)

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

1/(t-2)(t+2) = A/(t-2) +
B/(t+2)

1 = t(A+B) + 2A -
2B

A+B = 0

A-B =
1/2

-2B = 1/2

B = -1/4
=> A = 1/4

1/(t-2)(t+2) = 1/4(t-2) -
1/4(t+2)

Int dt/(t-2)(t+2) = Int dt/4(t-2) - Int
dt/4(t+2)

Int dt/(t-2)(t+2) = (1/4)*[ln|t-2| -
ln|t+2|]

We'll apply quotient rule of
logarithms:

Int dt/(t-2)(t+2) =
(1/4)*[ln|t-2|/|t+2|]

F(1) = (1/4)*[ln|1-2|/|1+2|] =
(1/4)*ln (1/3)

F(0) = (1/4)*[ln|0-2|/|0+2|] = (1/4)*ln 1 =
0

Int dt/(t-2)(t+2) = F(1) -
F(0)

Int dt/(t-2)(t+2) = (1/4)*ln
(1/3)

The area of the region located under
the curve f(x) and between x axis and the line x = 0 and x = pi/2,is A = (1/4)*ln (1/3)
square units.