Calculus of a Single Variable, Chapter 5, 5.7, Section 5.7, Problem 39

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