Area between a curve and two lines.Determine the area  between the curve y=cos x/(4+sin x), the lines x=0 and x=pi/2 and the x axis?

2 Answers | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to evaluate the area of the region between the  given curves, hence, you need to evaluate the definite integral, such that:

`int_0^(pi/2) cos x/(4 + sin x)dx`

You should come up with the substitution, such that:

`tan (x/2) = u => (1 + tan^2(x/2))/2 dx = du => dx = (2 du)/(1 + u^2)`

You need to express `sin x` and `cos x` in terms of `tan(x/2)` , such that:

`sin x = (2 tan(x/2))/(1 + tan^2(x/2))`

`cos x = (1 - tan^2(x/2))/(1 + tan^2(x/2))`

Changing the limits of integration yields:

`x = 0 => tan 0 = 0`

`x = pi/2 => tan(pi/4) = 1`

Changing the variable yields:

`int_0^1 ((1- u^2)/(1 + u^2))/((4 + (2u)/(1 + u^2)))*(2 du)/(1 + u^2)`  = `int_0^1(1 - u^2)/(4u^2 + 2u + 4)*(2 du)/(1 + u^2)`

You should use partial fractions decomposition, such that:

`(1 - u^2)/((2u^2 + u + 2)(1 + u^2)) = (Au + B)/(2u^2 + u + 2)+ (Cu + D)/(1 + u^2)`

`1 - u^2 = Au + Au^3 + B + Bu^2 + 2Cu^3 + Cu^2 + 2Cu + 2Du^2 + Du + 2D`

Equating coefficients of like powers yields:

`A + 2C = 0`

`B + C + 2D = -1`

`A + 2C + D = 0`

`B + 2D = 1 => C = -2 => A = 4 => D = 0, B = 1`

`(1 - u^2)/((2u^2 + u + 2)(1 + u^2)) = (4u + 1)/(2u^2 + u + 2)+ (-2u)/(1 + u^2)`

`int_0^1 (1 - u^2)/((2u^2 + u + 2)(1 + u^2))du = int_0^1 (4u + 1)/(2u^2 + u + 2) du - int_0^1 (2u)/(1 + u^2) du`

You need to evaluate `int_0^1 (4u + 1)/(2u^2 + u + 2) du` using the following substitution, such that:

`2u^2 + u + 2 = v => (u + 1)du = dv`

`int_0^1 (4u + 1)/(2u^2 + u + 2) du = int_2^5 (dv)/v = ln v|_2^5`

Using the fundamental theorem of calculus yields:

`int_0^1 (4u + 1)/(2u^2 + u + 2) du = ln 5 - ln 2 = ln (5/2)`

You need to evaluate `int_0^1 (2u)/(1 + u^2) du` using the following substitution, such that:

`1 + u^2 = v => 2u du = dv`

`int_0^1 (2u)/(1 + u^2) du = int_1^2 (dv)/v = ln v|_1^2`

`int_0^1 (2u)/(1 + u^2) du = ln 2 - ln 1 = ln 2`

`int_0^1 (1 - u^2)/((2u^2 + u + 2)(1 + u^2))du = ln (5/2) - ln 2`

`int_0^1 (1 - u^2)/((2u^2 + u + 2)(1 + u^2))du = ln (5/4)`

Hence, evaluating the area of the region bounded by the given curves, yields `int_0^(pi/2) cos x/(4 + sin x)dx = ln (5/4) .`

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.

We’ve answered 318,957 questions. We can answer yours, too.

Ask a question