The area bound by the curve y = cosx/(sin^2x-4), the x axis and the lines x = 0 and x = pi/2 is the integral of y between the limits x = 0 and x = pi/2.

y = cos x/ ((sin x)^2 - 4)

let sin x = y => dy = cos x dx

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

=> Int [ 1/ (y^2 - 4) dy]

=> Int [ 1/4*(y - 2) - 1/4*(y + 2) dy]

=> ln |y - 2|/4 - ln |y + 2|/4 + C

substitute y = sin x

=> ln |sin x - 2|/4 - ln |sin x + 2|/4 + C

F(pi/2) = ln|1|/4 - ln 3 / 4 + C

F(0) = ln 2 / 4 - ln 2 / 4 + C

F(pi/2) - F(0)

=> (- ln 3) / 4

Area cannot be negative, so we use the properties of logarithm and write

- ln 3 / 4 = [ln(1/3)]/4

**The required area is [ln(1/3)]/4**

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