Evaluate: $\int \sqrt {\frac {x}{a-x}}\, dx $

Question

Evaluate: $\int \sqrt {\dfrac {x}{a-x}}\, dx$

My Attempt: $$\int \sqrt {\dfrac {x}{a-x}} \, dx$$

Let $x=a\cos (2\theta)$ $$dx=-2a\sin (2\theta ) d\theta$$ Then, $$=-\int \sqrt {\dfrac {a \cos (2\theta)}{a-a\cos (2\theta )}} 2a\sin (2\theta ) d\theta$$ $$=-\int \sqrt {\dfrac {\cos (2\theta)}{1-\cos (2\theta)}} 2a\sin (2\theta) d\theta$$

Answer 1

I feel a better substitution would be: $$x =a\cos^2 \theta \implies dx =-a\sin 2\theta \, d\theta$$ giving us: $$I = \int \sqrt{\frac{x} {a-x}} \, dx$$ $$=-a\int \cot \theta \sin 2\theta\, d\theta$$ $$=-2a\int \cos^2 \theta \, d\theta$$

Hope you can take it from here.

Answer 2

Hint: Do the substitution $x=\dfrac{ay^2}{1+y^2}$ and $\mathrm dx=\dfrac{2ay}{(1+y^2)^2}\,\mathrm dy$. With this substitution,$$\sqrt{\frac x{a-x}}=y.\tag1$$In fact, that's where my substitution comes from, since$$(1)\iff\frac x{a-x}=y^2\iff x=\dfrac{ay^2}{1+y^2}.$$