Calculus of a Single Variable, Chapter 8, 8.4, Section 8.4, Problem 38

Recall that indefinite integral follows int f(x) dx = F(x) +C
where:
f(x) as the integrand function
F(x) as the antiderivative of f(x)
C as the constant of integration..
To evaluate the given integral problem: int x^2/sqrt(2x-x^2) dx , we apply completing the square on the expression: 2x-x^2 .
Completing the square:
Factor out (-1) from 2x-x^2 to get (-1)(x^2-2x)
The x^2-2x or x^2-2x+0 resembles ax^2+bx+c where:
a= 1 and b =-2 that we can plug-into (-b/(2a))^2 .
(-b/(2a))^2= (-(-2)/(2*1))^2
= (2/2)^2
= 1^2
=1
To complete the square, we add and subtract 1 inside the ():
(-1)(x^2-2x) =(-1)(x^2-2x+1 -1)
Distribute (-1) in "-1 "to move it outside the () .
(-1)(x^2-2x+1 -1)= (-1)(x^2-2x+1)+ (-1)(-1)
= (-1)(x^2-2x+1)+ 1
Apply factoring for the perfect square trinomial: x^2-2x+1= (x-1)^2
(-1)(x^2-2x+1)+ 1=-(x-1)^2 + 1
= 1-(x-1)^2
Apply 2x-x^2=1-(x-2)^2 to the integral, we get: int x^2/sqrt(1-(x-1)^2) dx
Apply u-substitution by letting u =x-1 then x = u+1 and du =dx . The integral becomes:
int x^2/sqrt(1-(x-1)^2) dx=int (u+1)^2/sqrt(1-u^2) du
Apply FOIL method on (u+1)^2 , we get:
(u+1)^2 = (u+1) *(u+1)
= u*u +u*1 + 1*u +1*1
= u^2 +u+u+1
= u^2+2u +1
Plug-in (u+1)^2= u^2+2u +1 on the integral, we get:
int (u+1)^2/sqrt(1-u^2) dx =int (u^2+2u +1)/sqrt(1-u^2) du
Apply the basic integration property: int (u+v+w) dx = int (u) dx + int (v) dx+int (w) dx .
int (u^2+2u +1)/sqrt(1-u^2) du=int u^2/sqrt(1-u^2) du +int (2u)/sqrt(1-u^2) du+int 1/sqrt(1-u^2) du
Each integral resembles formula from integration table for rational function with roots. For the first integral, we follow: int (x^2 dx)/sqrt(a^2-x^2) =-(xsqrt(a^2-x^2))/2 +(a^2arcsin(x/a))/2 +C .
Then,
int u^2/sqrt(1-u^2) du =-(usqrt(1-u^2))/2 +(1arcsin(u/1))/2
=-(usqrt(1-u^2))/2 +arcsin(u)/2
For second integral, we follow: int x/sqrt(a^2-x^2)dx= -sqrt(a^2-x^2)+C .
int (2u)/sqrt(1-u^2) du =2int u/sqrt(1-u^2) du
=2 *[-sqrt(1-u^2)]
=-2sqrt(1-u^2)
For the third integral, we follow: int dx/(a^2-x^2)dx=arcsin(x/a)+C .
int 1/sqrt(1-u^2) du = arcsin(u/1) or arcsin(u)
Combining the results, we get:
int (u^2+2u +1)/sqrt(1-u^2) du=-(usqrt(1-u^2))/2 +arcsin(u)/2-2sqrt(1-u^2)+arcsin(u) +C
Plug-in u = x-1 , we get the indefinite integral as:
int x^2/sqrt(2x-x^2) dx
=-((x-1)sqrt(1-(x-1)^2))/2 +arcsin(x-1)/2-2sqrt(1-(x-1)^2)+arcsin(x-1) +C
Recall 1-(x-1)^2 = 2x-x^2 then the integral becomes:
int x^2/sqrt(2x-x^2)dx
= (( -x+1)sqrt(2x-x^2))/2 +arcsin(x-1)/2-(4sqrt(2x-x^2))/2+(2arcsin(x-1))/2 +C
= [arcsin(x-1) + 2arcsin(x-1)]/2 + [( -x+1)sqrt(2x-x^2)-4sqrt(2x-x^2)]/2+C
=(3arcsin(x-1))/2 + ((-x-3)sqrt(2x-x^2))/2 +C
=(3arcsin(x-1))/2 +((-1)(x+3)sqrt(2x-x^2))/2 +C
=(3arcsin(x-1))/2-((x+3)sqrt(2x-x^2))/2 +C
or (3arcsin(x-1))/2-(xsqrt(2x-x^2))/2 -(3sqrt(2x-x^2))/2+C