Question

How do you find the integral of `int 1/(sqrtxsqrt(1-x)`?

Answer 1

`2arcsin(sqrtx)+C`

Explanation:

`I=int1/(sqrtxsqrt(1-(sqrtx)^2))dx`

Let `u=sqrtx` so `du=1/(2sqrtx)dx`.

`I=2int1/(2sqrtxsqrt(1-(sqrtx)^2))dx`

`I=2int1/sqrt(1-u^2)du`

You may recognize this as the arcsine integral, but we can substitute `u=sintheta` to show this. We also see that `du=costhetad theta`.

`I=2int1/sqrt(1-sin^2theta)(costhetad theta)`

Since `1-sin^2theta=cos^2theta`:

`I=2int1/costhetacosthetad theta`

`I=2intd theta`

`I=2theta+C`

From `u=sintheta` we see that `theta=arcsin(u)`:

`I=2arcsin(u)+C`

`I=2arcsin(sqrtx)+C`