# How to integrate $\int\frac{\sqrt{1-x}}{\sqrt{x}}\ \mathrm dx$

I'm having a bit of trouble solving this integral: $$\int\frac{\sqrt{1-x}}{\sqrt{x}}dx$$

Here is my attempt at a solution:

I multiplied the numerator and the denominator of $\frac{\sqrt{1-x}}{\sqrt{x}}$ by $\sqrt{x}$, yielding $$\int\frac{\sqrt{x-x^x}}{x}dx.$$ Further simplification resulted in $$\int\frac{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}{x}dx.$$ Using trigonometric substitution, I set $$x-\frac{1}{2}=\frac{1}{2}\sin\theta$$ and solving for the differential $dx$ got $$dx=\frac{1}{2}\cos\theta.$$ Substituting this all back into $\int\frac{\sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}}{x}dx$ (and some simplification later) yielded $$\frac{1}{2}\int{\frac{\cos^2\theta}{\sin\theta+1}}d\theta.$$ By substituting $1-\sin^2\theta$ for $\cos^2\theta$ I obtained $$\frac{1}{2}\int{\frac{1}{\sin\theta+1}-\frac{\sin^2\theta}{\sin\theta+1}d\theta}.$$ The issue I'm having is trying to solve this resultant integral. If there is an easier method to solve the problem, that would be graciously accepted.

Set $\sqrt{x} = \sin(t)$. We then have $x = \sin^2(t)$. Hence, $1-x = \cos^2(t)$. This gives us \begin{align} \int \dfrac{\sqrt{1-x}}{\sqrt{x}}dx & = \int \dfrac{\cos(t)}{\sin(t)} 2 \sin(t) \cos(t) dt = 2\int \cos^2(t) dt\\ & = \int(1+\cos(2t))dt = t + \dfrac{\sin(2t)}2 + c\\ & = \arcsin(\sqrt{x}) + \sqrt{x}\sqrt{1-x} + c \end{align}

• As $\sqrt{1-x}=|\cos t,|$ should the assumption that $\cos t\ge0$ not be included? – lab bhattacharjee Nov 30 '13 at 5:04
• @labbhattacharjee Not necessary. Since $\sqrt{x} > 0$, I restrict my $t$ from $0$ to $\pi/2$. – user17762 Dec 1 '13 at 5:03
• even that conclusion should be explicit :) – lab bhattacharjee Dec 1 '13 at 5:08

Let $$u=\frac{\sqrt{1-x}}{\sqrt{x}}$$

Then $u^2=\frac{1-x}{x}=\frac{1}{x}-1$. Hence $$x=\frac{1}{u^2-1}=\frac{1}{2(u-1)}-\frac{1}{2(u+1)}$$ $$dx=-\frac{1}{2(u-1)^2}+\frac{1}{2(u+1)^2}$$

$$\frac{1}{2} \int \frac{u}{(u+1)^2}-\frac{u}{(u-1)^2}du$$

Alternative solution, making the substitution $x=t^2$ integral becomes :

$$I=\int\frac{\sqrt{1-x}}{\sqrt{x}}\;\mathrm{d}x = 2\int\sqrt{1-t^2}\;\mathrm{d}t =2J$$

But that means :

$$J=\int\sqrt{1-t^2}\;\mathrm{d}t = \int\frac{1-t^2}{\sqrt{1-t^2}}\;\mathrm{d}t = \arcsin t - \int\frac{t^2}{\sqrt{1-t^2}}\;\mathrm{d}t =$$ ... via per partes ...

$$= \arcsin t + t\sqrt{1-t^2}-\int\sqrt{1-t^2}\;\mathrm{d}t = \arcsin t + t\sqrt{1-t^2} -J$$

Therefore

$$I=2J =\arcsin t + t\sqrt{1-t^2} = \arcsin \sqrt{x} + \sqrt{x}\sqrt{1-x}$$

However, your original way is not bad after all, if you continued - see :

$$\frac{1}{2}\int\frac{\cos^2{\theta}}{1+\sin\theta}\;\mathrm{d}\theta = \frac{1}{2}\int\frac{1-\sin^2{\theta}}{1+\sin\theta}\;\mathrm{d}\theta = \frac{1}{2}\int\frac{(1-\sin{\theta})(1+\sin{\theta})}{1+\sin\theta}\;\mathrm{d}\theta = \frac{1}{2}\int1-\sin{\theta}\;\mathrm{d}\theta$$

Therefore

$$I=\frac{1}{2}\theta+\frac{1}{2}\cos{\theta}=\frac{1}{2}\arcsin{(2x-1)}+\frac{1}{2}\sqrt{1-(2x-1)^2}= \frac{1}{2}\arcsin{(2x-1)}+\sqrt{x^2-x}$$

and these results are indeed equivalent, because

$$\frac{1}{2}\arcsin{(2x-1)}=\arcsin\sqrt{x}-\frac{\pi}{4}$$ multiplying by $2$ and taking sin of both sides :

$$2x-1=-\cos\left( 2\arcsin\sqrt{x}\right)=2\sin^2\arcsin\sqrt{x}-1=2x-1$$

Or let $\theta=\alpha-\pi/2$, then

$$x=\frac{1+\sin\theta}{2}=\frac{1-\cos\alpha}{2}=\sin^2\left(\frac{\alpha}{2}\right)$$

So

$$\frac{\theta}{2}=\frac{\alpha}{2}-\frac{\pi}{4}=\arcsin\sqrt{x}-\frac{\pi}{4}$$