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

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justaguide | College Teacher | (Level 2) Distinguished Educator

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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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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.

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