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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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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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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