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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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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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neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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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