# Why is $\int \frac{1}{\sqrt{y}\sqrt{1 - y}} dy = \frac{2\sqrt{y - 1}\sqrt{y} \log(\sqrt{y - 1} + \sqrt{y})}{\sqrt{(-(y - 1) y)}}$?

Fairly self-explanatory question title. Why is $$\int \frac{1}{\sqrt{y}\sqrt{1 - y}} dy = \frac{2\sqrt{y - 1}\sqrt{y} \log(\sqrt{y - 1} + \sqrt{y})}{\sqrt{-(y - 1)}\sqrt{y}}\ ?$$

I'm assuming you have to use substition, but I'm not sure how.

edit: $$y \in (0,1)$$

• try using substitution $y=t^2$ – Abhash Jha Apr 22 '17 at 13:46
• Substitute $y = \sin^2 \theta$. – user384138 Apr 22 '17 at 13:47
• constrain your allowed values for $y$ to $(0,1)$ and everything should become much clearer – tired Apr 22 '17 at 13:47
• The term $\frac{2\sqrt{y-1}\sqrt{y}}{\sqrt{-(y-1)y}}$ can be greatly simplified and I suggest to do it. – Jack D'Aurizio Apr 22 '17 at 13:55
• Please check the signs in your expression. The denominator on the LHS is defined for $0 < y < 1$, but the first term in the numerator on the RHS is defined for $y>1$. – mlc Apr 22 '17 at 14:03

Is the answer correct? Notice $y\in (0,1)$, as both $y> 0$, $1-y>0$. But why answer has $\sqrt{y-1}$?

Let $y=\sin ^2x$, $x\in (0,\frac{\pi}{2})$ $$\int \frac{1}{\sqrt{y}\sqrt{1 - y}} dy = \int \frac{2\sin x \cos x}{\sin x\cos x} dx =2x +C=2\arcsin \sqrt{y}+C$$

• The two different representations follow from $\arcsin(x) = i\log(\sqrt{1-x^2}+ix)$ – Hyperplane Apr 22 '17 at 14:20
• @Hyperplane yes, it we allow complex number, we could certainly find the relationship between them. But I did not go further steps there, because the question's tag does not involve complex analysis. – Yujie Zha Apr 22 '17 at 15:11
• You're right. The problem is indeed supposed to happen within the real numbers. Which means my setup isn't quite consistent. – ghthorpe Apr 22 '17 at 15:51

Hint:

As $$4y(1-y)=1-(2y-1)^2$$

Set $2y-1=\sin t$

Another possibility is to write it as $$\int\frac{\sqrt{y}}{y\sqrt{1-y}}dy$$and substitute $$u=\sqrt{\frac{y}{1-y}}$$ this avoids trigonometry.