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}

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

Your integral becomes

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


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 ...

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


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

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

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


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

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





Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.