Evaluating $\int^{x_2} _{x_1} \sqrt{a - b x^m} ~dx $

Question

Is there any way to evaluate $$\int^{x_2} _{x_1} \sqrt{(a - b x^m)}~ dx $$ where $x_{12} = \pm (a/b)^{1/m}$ without elliptic functions or hypergeometry? Or just any way to solve it. My attempt is to substitute $x = (a/b)^{1/m} \sin u$, $dx = (a/b)^{1/m} \cos u ~du $, $u = \pm \frac{\pi}{2}$. The initial expression then equals $$\int^{\frac{\pi}{2}} _{-\frac{\pi}{2}} \Big (\frac{a}{b}\Big)^{\frac{1}{m}} \sqrt{a (1-\sin^m u)}~ dx $$ It worked good before, when $m = 2$, but now it does not seem do help. Thank you for your answer.

Answer 1

Try substituting $x=(a/b)^{1/m} y^{1/m}$. This should reduce it to a multiple of an integral of the form $$ \int_{A}^B y^{k} (1-y)^n \, dy, $$ which can be at least partially computed using the Beta function.