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I'm not being able to calculate $ \large{\int{\sqrt{\frac{x}{a-x}}} dx} $ , someone could help me? I tryed to use integration by parts, but i achieved $0 = 0$.

Thanks in advance.

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It's messy... And scary. Wolfram: $$\frac{\sqrt{\frac{x}{x-a}}\left(\sqrt{x}(x-a)+a\sqrt{a-x}\tan^{-1}\left(\sqrt{‌​\frac{x}{a-x}}\right)\right)}{\sqrt{x}}$$ – Shahar May 16 '14 at 23:15
The components are very simple, the substitution $x=a\sin^2\theta$ (if $a\gt 0$) is natural, and then everything collapses. Wolfram makes things look harder than they really are. – André Nicolas May 16 '14 at 23:20
up vote 2 down vote accepted

First to make life a little bit easier on yourself, set $x = ay$ to turn your integral into

$$\int \sqrt{\frac{ay}{a-ay}} a\,dy = a\int\sqrt{\frac{y}{1-y}}\,dy.$$

Let's make the substitution $y = \cos^2(t)$. Then $dy = -2\sin(t)\cos(t)dt$ which gives

$$a\int \sqrt{\frac{\cos^2(t)}{1-\cos^2(t)}}(-2\sin(t)\cos(t))dt = -2a\int \sqrt{\frac{\cos^2(t)}{\sin^2(t)}}\sin(t)\cos(t)dt.$$

This integral is pretty manageable at this point. Can you take it from here?

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Note, you can pick $t$ so that $\sin t$ and $\cos t$ are positive... – Thomas Andrews May 16 '14 at 23:24
if i consider that, i can conclude that the element inside the integral is $\mid{\frac{\cos{t}}{\sin{t}} }\sin{t}\cos{t}$, but i cannot simply cut the \sin{t}, right? – Mike May 16 '14 at 23:25
@ExampleMo Look at Thomas' comment just above yours. – Cameron Williams May 16 '14 at 23:26
Ok, so i have $\int{\cos^2{t}}dt = \int{ \frac{1+ \cos{2t}}{2}}dt$,and i know how to solve. after that, i need to change back to my solution? – Mike May 16 '14 at 23:37
+1, for a short way you can set $x=a\cos^2t$. – Tunk-Fey May 17 '14 at 3:01

You could also substitute $u=\sqrt{\frac{x}{a-x}}$, $\;x=\frac{au^2}{1+u^2}$ and $dx=\frac{2au}{(1+u^2)^2}du$ to get $\int\frac{2au^2}{(1+u^2)^2}du$.

Then letting $u=\tan\theta$, $du=\sec^{2}\theta \;d\theta$ gives

$2a\int\sin^{2}\theta\; d\theta = a[\theta-\sin\theta\cos\theta]+C=a[\tan^{-1}u-\frac{u}{1+u^2}]+c$, and then substitute back for u.

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Let $\displaystyle u^2=\frac{x}{a-x}$. Then $x=\displaystyle\frac{au^2}{1+u^2}$ and so $\displaystyle du=\frac{2au\, du}{(1+u^2)^2}$. Then, $$\int \sqrt{\frac{x}{a-x}}dx = \int \sqrt{u^2}\frac{2au}{(1+u^2)^2}du=\int \frac{2au^2\, du}{(1+u^2)^2}. $$ Now, let $u=\tan t$ and $\displaystyle\cos t=\frac{1}{\sqrt{1+u^2}}$. Then $du=\sec^2 t\,dt$ and we have $$ \int \sqrt{\frac{x}{a-x}}dx =\int \frac{2au^2\, du}{(1+u^2)^2}=\int \frac{2a\tan^2 t\sec^2 t\, dt}{(\sec^2t)^2}=2a\int \frac{\tan^2t}{\sec^2t}dt=$$ $$=2a\int \frac{\sin^2t}{\cos^2t}\cdot \cos^2t\, dt=2a\int \sin^2t dt=2a\int \frac{1-\cos 2t}{2}dt= $$ $$=a\int dt-a\int \cos 2t \, dt = at-a\frac{1}{2}\sin 2t +c= $$ $$=at-a\sin t\cdot \cos t +c = a\arctan u-a\frac{u}{\sqrt{1+u^2}}\cdot \frac{1}{\sqrt{1+u^2}}+c= $$ $$= a\arctan u -\frac{au}{1+u^2}+c =$$ $$=a\arctan\sqrt{\frac{x}{a-x}}-\frac{\sqrt{\frac{x}{a-x}}}{1+\frac{x}{a-x}} +c.$$

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Integrals of roots (of any order) of a linear function can be integrated by substitution, in this case of $$y^2=\frac{x}{a-x}$$ solving for $x$ we get $$x=\frac{ay^2}{y^2+1}$$ so $$dx=\frac{2ay}{(y^2+1)^2}dy$$ substituting this gives, $$\int\sqrt{\frac{x}{a-x}}dx=\int \frac{2ay^2}{(y^2+1)^2}dy$$

which is a rational function and can be integrated by partial fractions. In this case $$\int \frac{y^2}{(y^2+1)^2}dy=\int \frac{1}{y^2+1}dy-\int \frac{1}{(y^2+1)^2}dy$$

$$\int \frac{1}{y^2+1}dy=\arctan(y)$$ is easy but the other integral is harder and can be done in several ways including trig substitution, but can also be done directly by parts \begin{equation*} \begin{split} \int \frac{1}{y^2+1}dy&=\frac{y}{y^2+1}-\int \frac{2y^2}{(y^2+1)^2}dy\\ &=\frac{y}{y^2+1}-2\int \frac{1}{y^2+1}dy +2\int \frac{1}{(y^2+1)^2}dy\\ \end{split} \end{equation*}

So \begin{equation*} \begin{split} 2\int \frac{1}{(y^2+1)^2}dy&=3\int \frac{1}{y^2+1}dy -\frac{y}{y^2+1}\\ &=3\int \arctan(y) -\frac{y}{y^2+1}\\ \end{split} \end{equation*}

And we have

$$\int \frac{y^2}{(y^2+1)^2}dy= -\frac{1}{2}\int \arctan(y) +\frac{1}{2}\frac{y}{y^2+1}$$

Then assuming I have not made any mistakes, (which is unlikely) you should be able to show it is equal to the Wolfram answer.

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