`int_2^3 (2x-3)/sqrt(4x-x^2) dx` Find or evaluate the integral by completing the square

Recall  that `int_a^b f(x) dx = F(x)|_a^b` :

`f(x)` as the integrand function

`F(x) ` as the antiderivative of `f(x)`

"a" as the lower boundary value of x

"b" as the upper boundary value of x

To evaluate the given problem: `int_2^3 (2x-3)/sqrt(4x-x^2)dx` , we need to determine the

 indefinite integral F(x)  of the integrand: `f(x)=(2x-3)/sqrt(4x-x^2)` .

We apply completing the square on `4x-x^2` .

Factor out `(-1)`  from `4x-x^2` to get `(-1)(x^2-4x)`

The `x^2-4x` or `x^2-4x+0` resembles `ax^2+bx+c` where:

`a= 1` and `b =-4` that we can plug-into `(-b/(2a))^2` .

`(-b/(2a))^2= (-(-4)/(2*1))^2`

              `= (4/2)^2`

              ` = 2^2`

              ` =4`

To complete the square, we add and subtract 4 inside the ():

`(-1)(x^2-4x) =(-1)(x^2-4x+4 -4)`

Distribute (-1) in "-4" to move it outside the ().

`(-1)(x^2-4x+4 -4) =(-1)(x^2-4x+4) + (-1)(-4)`

                          `=(-1)(x^2-4x+4) + 4`

Apply factoring for the perfect square trinomial: `x^2-4x+4 = (x-2)^2`

`(-1)(x^2-4x+4) + 4 =-(x-2)^2 + 4`

                                     ` = 4-(x-2)^2`

which means `4x-x^2=4-(x-2)^2`

Applying it to the integral:

`int_2^3 (2x-3)/sqrt(4x-x^2)dx =int_2^3 (2x-3)/sqrt(4-(x-2)^2)dx`

To solve for the indefinite integral of `int (2x-3)/sqrt(4-(x-2)^2)du` ,

let `u =x-2` then `x = u+2` and `du= dx` .

Apply u-substitution , we get:

`int (2x-3)/sqrt(4-(x-2)^2)dx= int (2(u+2)-3)/sqrt(4-u^2)du`

                             `=int (2u+4-3)/sqrt(4-u^2)du`

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

 Apply the basic integration property: `int (u+v) dx = int (u) dx + int (v) dx` .  

  `int (2u+1)/sqrt(4-u^2)du =int (2u)/sqrt(4-u^2)du +int1/sqrt(4-u^2)du`

For the integration of the first term: `int (2u)/sqrt(4-u^2)du` ,

let `v = 4-u^2` then `dv = -2u du` or `-dv = 2u du` then it becomes:

`int (2u)/sqrt(4-u^2)du =int (-1)/sqrt(v)dv`

Applying radical property: `sqrt(x) = x^(1/2)` and  Law of exponent: `1/x^n = x^-n` , we get:

`(-1)/sqrt(v) =(-1)/v^(1/2)`

Then,

`int (-1)/sqrt(v)dv =int(-1)v^(-1/2) dv`

Applying Power Rule of integration: `int x^n dx = x^(n+1)/(n+1)`

`int (-1)v^(-1/2) dv = (-1)v^(-1/2+1)/(-1/2+1)`

                         `=(-1)v^(1/2)/(1/2)`

                        `=(-1)v^(1/2)*(2/1)`

                         `=-2v^(1/2)`

                         `= -2sqrt(v)`

Recall `v =4-u^2 then-2sqrt(v)=-2sqrt(4-u^2)` .

Then,

`int (2u)/sqrt(4-u^2)du =-2sqrt(4-u^2)`

For the integration of the second term:  `int1/sqrt(4-u^2)du` ,

 we apply the basic integration formula for inverse sine function:

`int 1/sqrt(a^2-u^2) du = arcsin(u/a)`

Then,

`int1/sqrt(4-u^2)du=int1/sqrt(2^2-u^2)du`

                    `= arcsin(u/2)`

 For the complete indefinite integral, we combine the results as:

`int (2u+1)/sqrt(4-u^2)du =-2sqrt(4-u^2) +arcsin(u/2)`

 Then plug-in `u=x-2` to express it terms of x, to solve for `F(x)` .

`F(x) =-2sqrt(4-(x-2)^2) +arcsin((x-2)/2)`

For the definite integral, we applying the boundary values: `a=2` and `b=3` in `F(x)|_a^b= F(b) - F(a)` .

`F(3) -F(2) = [-2sqrt(4-(3-2)^2) +arcsin((3-2)/2)] -[-2sqrt(4-(2-2)^2) +arcsin((2-2)/2)]`

       `=[-2sqrt(4-(1)^2) +arcsin(1/2)] -[-2sqrt(4-(0)^2) +arcsin(0/2)]`

        `=[-2sqrt(3) +arcsin(1/2)] -[-2sqrt(4) +arcsin(0)]`

         ` =[-2sqrt(3) +pi/6] -[-2*(2)+0]`

        `=[-2sqrt(3) +pi/6] -[-4]`

         `=-2sqrt(3) +pi/6 + 4`