Integral in physics - how to evaluate it?

Question

This comes from a physics problem. For $r\le R$, consider $$t=\sqrt{\frac{R}{2c}} \int_r^R \sqrt{\frac{x}{R-x}} dx$$

The indefinite integral is (by the Wolfram Alpha): $$\frac{\sqrt{-\frac{x}{x-R}}(\sqrt{x}(x-R)+R\sqrt{R-x}\tan^{-1}{\frac{\sqrt{r}}{\sqrt{R-x}}}}{\sqrt{x}}$$ But it did not show the steps. How could I get the indefinite integral?

It's not a homework. I could finish the problem without knowing the method, but I'd like to know how to do this.

Answer 1

HINT...try substituting $$x=R\cos^2\theta$$

Answer 2

WLOG, $R=1$.

Complement the argument and get, with $r=1-s=1-t^2$,

$$\int\sqrt{\frac{1-s}{s}}ds=2\int\sqrt{1-t^2}dt.$$

This is the well-known integral for the area of a circle, which is the algebraic sum of a sector and a triangle,

$$\frac12\arcsin t+\frac12 t\sqrt{1-t^2}.$$

Notice that $\arcsin t=\arctan\dfrac t{\sqrt{1-t^2}}$.

Answer 3

Hint: successively substitute $y = R - x$ followed by $\frac Ry = \sec^2 \theta$. The final integral is easy to evaluate using a double angle cosine identity.