Ask a New Question
Menu

I have to find the integral of 1/(sq. rt. of (4x-x²)) dx. I know I have to complete the square of the denominator, but im confused on how that process works when im only given 4x-x². Could you demonstrate how to complete the square for that equation? Thank you so much!

1 answer

4x - x^2
-(x^2 - 4x)
x^2 - 4x = 0
x^2 - 4x + 4 = 0 + 4
(x - 2)^2 = 4
(x - 2)^2 - 4
-(x - 2)^2 + 4
4 - (x - 2)^2

| = integral sign
| 1/(sqrt(4 - (x - 2)^2) dx

u = ( x - 2)
du = dx

| 1/(sqrt(4 - u^2)) du
= arcsin u/2 + C
= arcsin (x - 2)/2 + C
Similar Questions
  1. find integral using table of integrals )integral sin^4xdx this the formula i used integral sin^n xdx =-1/n sin^n-1xcosx +n-1/n
    1. answers icon 3 answers
  2. LEt f and g be continous functions with the following propertiesi. g(x) = A-f(x) where A is a constant ii. for the integral of 1
    1. answers icon 0 answers
  3. LEt f and g be continous functions with the following propertiesi. g(x) = A-f(x) where A is a constant ii. for the integral of 1
    1. answers icon 0 answers
  4. LEt f and g be continous functions with the following propertiesi. g(x) = A-f(x) where A is a constant ii. for the integral of 1
    1. answers icon 1 answer
more similar questions