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

Question

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 $$\int\frac{\sqrt{x-x^2}}{x}dx.$$ 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 $$dx=\frac{1}{2}\cos\theta d\theta$$ 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.

Answer 1

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}

Answer 2

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

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

Answer 3

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

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

Your integral becomes

$$-2 \int \frac{u^2}{(u^2+1)^2}du=-2 \int \frac{u^2+1}{(u^2+1)^2}du+2 \int \frac{1}{(u^2+1)^2}du$$ which can be calculated integrating by parts $\frac{1}{u^2+1}$ or via a standard trig substitution.

Answer 4

Integrate by parts $$\int\frac{\sqrt{1-x}}{\sqrt{x}}dx =\sqrt x \sqrt{1-x}-\int \frac{d(\sqrt x)}{\sqrt{1-x}} = \sqrt{x(1-x)}-\sin^{-1}\sqrt x$$

Answer 5

Exponential substitution gives a general approach for such integrals. Applying the general transformation formula $(24)$, this is

$$ \boxed{ \int f\left( x, \sqrt{(x + b)^2 - a^2}, \dfrac{ \sqrt{x + b - a} }{ \sqrt{x + b + a} } \right) \, dx = \int f\left( \dfrac{e^{i\alpha} + e^{-i\alpha}}{2} a - b, \, \dfrac{ e^{\mp i\alpha} - e^{\pm i\alpha} }{2} a, \, \dfrac{1 - e^{\pm i\alpha}}{1 + e^{\pm i\alpha}} \right) \dfrac{ e^{-i\alpha} - e^{i\alpha} }{2i} a \, d\alpha } $$

Where: $\alpha = \cos^{-1}\left( \dfrac{x + b}{a} \right)$. For the alternating signs $\mp$ and $\pm$, use the upper sign when $\dfrac{x + b}{a} \geq 1$, and use the lower sign when $0 \leq \dfrac{x + b}{a} \leq 1$.

Since the integral is defined for $0 < x \leq 1$, we'll use the lower sign. The values of $a$ and $b$ are obtained by solving the following simple system of equations:

$$\left. b-a=-1\atop a+b=0\right\}$$

The solutions are $a = \frac12$ and $b = -\frac12$. Applying formula $(24)$ in the aforementioned link for $\frac12\leq x \leq1$, the integral becomes $$\begin{aligned}\int \frac{\sqrt{1-x}}{\sqrt{x}}\,dx&=i\int \frac{\sqrt{x-1}}{\sqrt{x}}\,dx\\&=-\frac{1}{4}\int (e^{-i\alpha}+e^{i\alpha}-2)\,d\alpha\\&=-\frac{ i e^{i \alpha} - i e^{-i \alpha} + 2 \alpha }{ 4 }+C\\&=\frac{ \sin \alpha - \alpha }{ 2 }+C\\&=\frac{ \sin \left( \cos^{-1}(2x - 1) \right) - \cos^{-1}(2x - 1) }{ 2 }+C\\&=\sqrt{x(1 - x)}\ -\ \dfrac{1}{2}\cos^{-1}(2x - 1)+C \end{aligned}$$

Another example solved by this method can be found here.

Answer 6

Integrate $\int \frac{\sqrt{1-x}}{\sqrt x} dx$

let $$ \begin{equation} \tag 1 x=t^2, \,\,\,\,\, dx=2t\, dt \end{equation} $$ now; $\int \frac{\sqrt{1-t^2}}{t} dt$ $$ \begin{equation} \tag{as $dt=\frac{dx}{2t}$} \int 2 \sqrt{1 - t^2} dt \end{equation} $$ $$ \begin{equation} \tag 2 2 \int \sqrt{1 - t^2} dt \end{equation} $$ $$ \begin{equation} \tag 3 2 \left[\frac{t}{2}\sqrt{1-t^2}+\frac{1}{2}\sin^{-1} (t) \right] \end{equation} $$

put $t=\sqrt{x}$, to get answer.