Evaluate $\int (1-x^2)^{1/4} \,dx$

Question

I came across this integral while originally evaluating $\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \,dx$ which led me to two integrals, one, $\sqrt{x}/(1-x^2)^{1/2}$ which on substituting $(1-x^2)^{1/2}$ as t gives $2(1-t^2)^{1/4}$, which is in the title as well and the other a simple $1/(1-x^2)^{1/2}$ which yeilds $\arcsin(x)$. I can't seem to find a method to solve that integral, would appreciate any help you can offer.

Answer 1

This integral doesn't appear to have a elementary antiderivative---or at least neither Maple nor SageMath could find one. Maple gives $$\int (1 - x^2)^{1 / 4} dx = x \cdot {}_2 F_1\left(-\frac{1}{4}, \frac{1}{2}; \frac{3}{2}; x^2\right) + C,$$ where ${}_2 F_1$ is the ordinary hypergeometric function.

Incidentally, to evaluate the original integral, $$\int \sqrt\frac{1 - \sqrt{x}}{1 + \sqrt{x}} \,dx ,$$ the substitution $x = u^2, dx = 2 u \,du$ transforms the integral into one with an integrand a rational function of $u$ and $\sqrt{1 - u^2}$, which can thus in turn be rationalized with an Euler substitution.

Answer 2

Note that: $$ \int \frac{\sqrt x}{\sqrt{1-x^2}}\,dx = \int (1-t^2)^{1/4}\frac{-2t}{2t\sqrt{1-t^2}}\,dt $$ But in fact you can work directly with what you originally have: $$ I= \int\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}\,dx =\int \sqrt{\frac{1-u}{1+u}}2u\,du = \int 4(z^2-1)\sqrt{2-z^2} \,dz$$

using the substitutions $x=u^2, u=z^2 -1$. Then, let $z=\sqrt{2}\sin{y}$ which gives you:

$$ I= \int 4(2\sin^2{y}-1)\sqrt{2}\cos{y},\ dy = 4\sqrt{2} \int \cos{y}\sin^2{y} -\cos^3{y} \,dz $$ $$= \frac{4\sqrt{2}}{3} \sin^3{y} - 4\sqrt{2}(\sin{y}-\frac{\sin^3{y}}{3}) + Constant $$

Finally you can then rewrite the result in terms of the variable $x$ using: $$ y=\arcsin{\frac{z}{\sqrt{2}}}=\arcsin{\frac{\sqrt{\sqrt{x}+1}}{\sqrt{2}}} $$