Determine the area between the curve y=cos x/(4+sin x), the lines x=0 and x=pi/2 and the x axis?
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First we find the integral of y=cos x/(4+sin x)
Let 4 + sin x = t
=> dt/dx = cos x
=> dt = cos x dx
Int [ cos x/(4+sin x) dx]
=> Int [ (1/t) dt]
=> ln t + C
=> ln ( 4 + sin x) + C
For x= 0
ln ( 4 + sin x) + C = ln (4 + 0) + C = ln 4 + C
For x = pi/2
ln ( 4 + sin x) + C = ln 5 + C
The difference of the integral for x = pi/2 and x = 0
=> ln 5 - ln 4
Therefore the required area is ln(5/4)
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To determine the area under y = cos x/(4+sin x). from x= 0 to x= pi/2.
We put 4+sinx = t. When x= 0, t = 4+sin0= 4.
When x= pi/2, t = 4+sinpi/2 = 4+1 =5.
Also by differentiating 4+six = t, we get cosx dx = dt.
So the given integral {f(x) dx from x= 0 to x= pi/2 }= dt/t from t= 4 to 5.
=( lnt +C) from t= t= 0 t o t = 5.
= ln5 - log4.
ln(5/4) = = log1.25.
Therefore Int {cosx/(4+sinx) dx from x= 0 to pi/2 } = ln 1.25.
We'll apply the Leibniz Newton formula to determine the area located between the given curve and lines.
Int f(x)dx = F(b) - F(a), where a = 0 and b = pi/2
We'll calculate the integral of f(x) = cos x/(4+sin x):
Int cos x dx/(4+sin x)
We'll substitute 4 + sin x = t
We'll differentiate both sides:
cos xdx = dt
We'll re-write the integral:
Int cos x dx/(4+sin x) = Int dt/t = ln |t| + C
We'll determine F(b) - F(a):
F(pi/2) = ln (4 + sin pi/2) = ln (4 + 1) = ln 5
F(0) = ln (4 + sin 0) = ln (4 + 0) = ln 4
Int f(x)dx = F(pi/2) - F(0)
Int f(x)dx = ln 5 - ln 4
Int f(x)dx = ln (5/4)
The area under the curve is ln (5/4) square units.
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