Hi I was wondering if someone would be able to help me evaluate \int \sqrt{x/1-x} with limits of 0 and 0.5, by using the substitution of x=sin2t - (thats sin squared t)
If x = sin^2(t), then dx/dt = 2sin(t)cos(t), or dx = 2sin(t)cos(t)dt.
So int{sqrt[x/(1 - x)]dx} = int{sqrt[sin^2(t)/(1 - sin^2(t))]2sin(t)cos(t)dt}
= int{2sqrt[sin^2(t)/cos^2(t)]sin(t)cos(t)dt}
= int{2[sin(t)/cos(t)]sin(t)cos(t)dt}
= int{2sin^2(t)dt}.
Can you go from here?