# Why are the two calculations of $\int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x$ give two distinct answers?

One of the calculations: \begin{aligned} \int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x &=\int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{1}{\sqrt{x}} \mathrm{~d} x \\ &=2 \int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \mathrm{~d}(\sqrt{x})=2 \arcsin \sqrt{x}+C . \end{aligned} The other: $$\int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x = \int \frac{d\left(x-\frac{1}{2}\right)}{\sqrt{\left(\frac{1}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}}} = \arcsin \frac{x-1 / 2}{1 / 2}+C=\arcsin(2x-1)+C$$

• The formula for $\cos{2u}=2\cos{u}^2-1$ can be used to get a relation between the two forms of arcsin appearing in the results and thus show both answers are equivalent. Jun 23, 2022 at 8:19
• There is no problem: the functions on the r.h.s. are the same except for an additive constant Jun 23, 2022 at 8:31

The answers are actually the same.

\begin{align*} \theta = 2\arcsin \sqrt x &\implies \sin \frac \theta 2 = \sqrt x \\ &\implies \sin^2 \frac \theta 2 = x \\ &\implies \frac{1-\cos \theta}{2} = x \\ &\implies \cos \theta = 1-2x \\ &\implies \sin(\theta+\pi/2) = 2x-1 \\ &\implies \theta = \arcsin(2x-1) - \pi/2 \end{align*}

And $$\pi/2$$ gets absorbed into $$+C$$

Because the integral is indefinite.

If you differentiate the first form, $${d2\arcsin{(\sqrt{x})} \over dx}=2{1 \over \sqrt{1-(\sqrt{x})^2}}\cdot {1 \over 2\sqrt{x}}={1 \over \sqrt{x(1-x)}}$$

As for the second one, $${d\arcsin{(2x-1)} \over dx}={1 \over \sqrt{1-(2x-1)^2}}\cdot 2={1 \over \sqrt{x(1-x)}}$$

Therefore, the both forms are correct.

They are both correct. Let $$y=2 \arcsin \sqrt x$$. Then $$x=(\sin (\frac y 2))^{2}$$. So $$2x-1=2(\sin (\frac y 2))^{2}-1=\sin (y+\frac {\pi} 2)$$. So $$y=\sin (2x-1)-\frac {\pi} 2$$.