I'm trying to find the antiderivative of the following function : $$f(x)= \left(1-x^2\right)^\frac{1}{m}$$ where $m \in \mathbb{N}$
I am looking for special cases for example $m=3$, $m=4$, $m=5$.
Please could you help me in finding how could I find the primitive of this function? Is there any particular technique concerning this types of functions?
Thanks in advance
Denoting $f(x) = \int_0^x \left(1 - t^2 \right)^{\frac{1}{m}} \, \mathrm{d}t$, by the FTC you know $f$ is an antiderivative you're looking for. Then $$ \int_0^x \left(1 - t^2 \right)^{\frac{1}{m}} \, \mathrm{d}t\overset{\color{purple}{u = \frac{t^2}{x^2}}}{=} \int_{0}^{1}\frac{1}{ \left(1 - x^2u \right)^{-\frac{1}{m}}} \frac{x}{2\sqrt{u}} \mathrm{d}u =x\frac{1}{B\left(\color{green}{\frac{1}{2}}, \color{maroon}{\frac{3}{2}}- \color{green}{\frac{1}{2}}\right)}\int_{0}^{1}\frac{u^{\color{green}{\frac{1}{2}} -1}(1-u)^{\color{maroon}{\frac{3}{2}} - \color{green}{\frac{1}{2}} -1}}{ \left(1 - \color{red}{x^2}u \right)^{\color{blue}{-\frac{1}{m}}}} \mathrm{d}u $$ since $B\left(\frac{1}{2}, \frac{3}{2}- \frac{1}{2}\right) = 2$ (with $B(x,y)$ being the Beta function). But recalling that the Hypergeometric function has integral representation $$ _2 F_1(\color{blue}{a},\color{green}{b};\color{maroon}{c};\color{red}{z}) = \frac{1}{B(\color{green}{b}, \color{maroon}{c}-\color{green}{b})} \int_0^1\frac{u^{\color{green}{b}-1}(1-u)^{\color{maroon}{c}-\color{green}{b}-1}}{(1-\color{red}{z}u)^{\color{blue}{a}}}\mathrm{d} u $$ you can conclude that $$ \boxed{\int\left(1 - x^2 \right)^{\frac{1}{m}} \, \mathrm{d}x = x\, _2F_1\left(-\frac{1}{m},\frac{1}{2};\frac{3}{2};x^2\right) +C} $$