Question

How do you evaluate?

`int1/(2*sqrt(x-x^2`dx

Answer 1

`int dx/(2sqrt(x-x^2)) = 1/2arcsin(2x-1)+C`

Explanation:

Complete the square at the denominator:

`int dx/(2sqrt(x-x^2)) = int dx /sqrt(1-1+4x-4x^2) = int dx /sqrt(1-(2x-1)^2)`

Substitute `sint = (2x-1)`, `x = (sint-1)/2`, `dx = 1/2 cost`, with `t in [-pi/2,pi/2]` so that `-1 <= 2x-1 <= 1` which is the interval in which the integrand is defined. In this interval `cost >=0`, so:

`int dx/(2sqrt(x-x^2)) = 1/2 int (costdt)/(sqrt(1-sin^2t))`

`int dx/(2sqrt(x-x^2)) = 1/2 int (costdt)/cost = 1/2 int dt=1/2 t + C`

and undoing the substitution:

`int dx/(2sqrt(x-x^2)) = 1/2arcsin(2x-1)+C`