Calculus of a Single Variable, Chapter 5, 5.7, Section 5.7, Problem 38

To evaluate the given integral: int 2/sqrt(-x^2+4x)dx , we may apply the basic integration property: int c*f(x)dx= c int f(x)dx .
The integral becomes:
2 int dx/sqrt(-x^2+4x)
We complete the square for the expression (-x^2+4x) .
Completing the square:
For the first step, factor out (-1): (-x^2+4x) = (-1)(x^2-4x) or -(x^2-4x)
The x^2 -4x or x^-4x+0 resembles the ax^2+bx+c where:
a=1 , b =-4 and c=0 .
To complete the square, we add and subtract (-b/(2a))^2 .
Using a=1 and b=-4 , we get:
(-b/(2a))^2 =(-(-4)/(2(1)))^2
=(4/2)^2
= 2^2
=4
Add and subtract 4 inside the (x^2-4x) :
-(x^2-4x+4 -4)
Distribute the negative sign on -4 to rewrite it as:
-(x^2-4x+4) +4
Factor the perfect square trinomial: x^2-4x+4 = (x-2)^2 .
-(x-2)^2 +4

For the original problem, we let: -x^2+4x=-(x-2)^2 +4 :
2 int dx/sqrt(-x^2+4x)=2 int dx/sqrt(-(x-2)^2+4)
It can also be rewritten as:
2 int dx/sqrt(-(x-2)^2 +2^2) =2 int dx/sqrt(2^2 -(x-2)^2)
The integral part resembles the integral formula:
int (du)/sqrt(a^2-u^2) = arcsin(u/a)+C .
Applying the formula, we get:
2 int dx/sqrt(2^2 -(x-2)^2) =2 *(arcsin (x-2)/2) +C
Then the indefinite integral :
int 2/sqrt(-x^2+4x)dx = 2arcsin((x-2)/2)+C