Evaluate the integral, for constants a>b: I = integral

Evaluate the integral, for constants $a>b$: $$ I = \int \sqrt{\frac{a-x}{x-b}}\,dx. $$

Step 1 · Substitute

Let $x=b+(a-b)\sin^2\theta$, $0\leq\theta\leq\tfrac\pi2$, so $dx=2(a-b)\sin\theta\cos\theta\,d\theta$, $x-b=(a-b)\sin^2\theta$, $a-x=(a-b)\cos^2\theta$:

$$ \sqrt{\frac{a-x}{x-b}} = \frac{\cos\theta}{\sin\theta} = \cot\theta \quad\Rightarrow\quad I = 2(a-b)\int\cos^2\theta\,d\theta. $$

Step 2 · Integrate

$$ 2(a-b)\int\cos^2\theta\,d\theta = (a-b)\int(1+\cos2\theta)\,d\theta = (a-b)\theta + (a-b)\sin\theta\cos\theta + C. $$

Step 3 · Convert back to $x$

Since $\sin\theta=\sqrt{(x-b)/(a-b)}$, $\cos\theta=\sqrt{(a-x)/(a-b)}$, $\theta=\arcsin\sqrt{(x-b)/(a-b)}$, and $(a-b)\sin\theta\cos\theta=\sqrt{(x-b)(a-x)}$:

$$ I = (a-b)\arcsin\sqrt{\frac{x-b}{a-b}} + \sqrt{(x-b)(a-x)} + C. $$

Summary Table

Additional Notes

• The domain is $b<x\leq a$.
• The result was verified numerically at $a=3,b=1,x=2$: the derivative evaluates to $1$, matching the integrand $\sqrt{(3-2)/(2-1)}=1$ there.
• Setting $a=2,b=0$ gives $\int\sqrt{(2-x)/x}\,dx=2\arcsin\sqrt{x/2}+\sqrt{x(2-x)}+C$ — note this is the reciprocal integrand of Problem 302 ($\sqrt{x/(2-x)}$ there, versus $\sqrt{(2-x)/x}$ here), so the two results are genuinely different integrals, not restatements of each other.