# Integral of $\int \frac{1}{\sqrt{x(1-x)}} dx$

$$\int \frac{1}{\sqrt{x(1-x)}} dx$$ I solved the integral in this way: make the substitution $x=\sin^2(u)$, then $dx=2\sin(u)\cos(u)du$. So the integral now becomes $$\int \frac{2\sin(u)\cos(u)}{\sqrt{\sin^{2}(u)(1-\sin^{2}(u))}} du=\int 2 du=2u+C.$$ Then subbing in $u=\arcsin(\sqrt{x})$ I get the solution $$2\arcsin(\sqrt{x})+C.$$ However when I typed in this integral onto Wolfram it gave me this.

So my question is did I get it wrong or are the two forms equivalent?

• These answers are equivalent for $0<x<1$. Jan 29, 2013 at 16:53
• Note that the Wolfram answer requires complex numbers in this range. There is a connection between the complex logarithm and the inverse trigonometric functions derived from the similar connection between the complex exponential and the trigonometric functions that explains this phenomenon. Jan 29, 2013 at 16:54
• FWIW, the graph of your solution matches the graph of the real part of the Wolfram solution Jan 29, 2013 at 17:03
• @Sasha totally forgot about the sqrt and the range of real values it could take and also the fact that Wolfram gives complex solutions. Jan 29, 2013 at 17:05

First of all, we never have to ask whether we got a correct antiderivative - because we can just take the derivative of our answer and see what we get: $$\frac d{dx} \big( 2\sin^{-1}(\sqrt x) +C \big) = \frac2{\sqrt{1-(\sqrt x)^2}}\frac d{dx}\sqrt x = \frac2{\sqrt{1-x}} \frac1{2\sqrt x} = \frac1{\sqrt{x(1-x)}}.$$ So you did it right.

There's another way of finding the antiderivative: complete the square inside the square root to see that $$\frac1{\sqrt{x(1-x)}} = \frac1{\sqrt{1/4 - (x-1/2)^2}} = \frac2{\sqrt{1-(2x-1)^2}}.$$ Therefore, using the substitution $u=2x-1$, \begin{align*} \int \frac1{\sqrt{x(1-x)}} \,dx = \int \frac2{\sqrt{1-(2x-1)^2}} \,dx &= \int \frac1{\sqrt{1-u^2}} \,du \\ &= \sin^{-1} u + C = \sin^{-1} (2x-1) + C. \end{align*} This can also be verified by differentiating.

(Side note: it's possible to check directly that $\sin^{-1}(2x-1)+\pi/2 = 2\sin^{-1}(\sqrt x)$, by taking the cosine of both sides, using $\cos(\theta+\pi/2) = -\sin\theta$ on the left and the double-angle formula $\cos 2\theta = 1-2\sin^2\theta$ on the right. Pretty cool!)

Put $\displaystyle x = \frac{1}{2}+t$ and $dx = dt$

So $\displaystyle \int\frac{1}{\sqrt{x.(1-x)}}dx$ is converted into

$\displaystyle \int\frac{1}{\sqrt{\frac{1}{2^2}-t^2}}dt = \sin^{-1}(2t)+\mathbb{C} = \sin^{-1}\left(2x-1\right)+\mathbb{C}$

Alternatively, let $$t = \sqrt{x}$$, then the integrand becomes $$\frac1{\sqrt{x(1-x)}} = \frac1{\sqrt x} \frac1{\sqrt{1-x}} = \frac1t \cdot \frac1{\sqrt{1-t^2}}$$

And $$dx = 2t \, dt$$. The integral becomes $$\int \frac1{t \sqrt{1-t^2}} \cdot 2t \, dt = 2 \int \dfrac1{\sqrt{1-t^2}} \, dt = 2 \arcsin(t) + C = 2\arcsin(\sqrt x) + C.$$