# Calc

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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!

• Calc -

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

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