Calculate the integral, with the help of Euler's integral Calculate the follwing integral:
$$\int_0^1 \sqrt{\frac{1-x^2}{x+2x^3+x^5}}dx$$
 A: Hint:
$$
x+2x^3+x^5 = x(1+x^2)^2
$$
Now find a substitution based on this that makes the integral look more like the Beta function.

Substitute $x=\sqrt{u}$, giving
$$
\int_0^1 \sqrt{\frac{1-u}{(1+u)^2}}\frac{du}{2u^{3/4}} = \frac12\int_0^1 \frac{\sqrt{1-u}}{(1+u)u^{3/4}}du
$$
You will notice that this is already quite close to being in the form of the Beta function. From here, you can use a taylor series on $\frac1{1+u}$ (which will converge correctly for all points in the integral) to get a sum of beta functions. This is how one would calculate it using Euler's Beta function.
It can be integrated directly to obtain a beta function, but this requires some more clever work. Let $\frac{1-x^2}{x}=2A^2$, so $x=\frac{\sqrt{A^4+4}-A^2}{2}$ and $\frac{dx}{1+x^2} = -\frac{2A}{A^4+4}dA$
$$
\int_0^1 \sqrt{\frac{1-x^2}{x}}\frac{dx}{1+x^2} = \sqrt{2}\int_0^\infty \frac{A^2}{A^4+1}dA
$$
Now, let $A^4=B$, so $A=\sqrt[4]{B}$ and $dA=\frac{dB}{4B^{3/4}}$, giving
$$
\sqrt{2}\int_0^\infty \frac{1}{B+1}\frac{dB}{4B^{1/4}}
$$
Now, let $B=\frac1C-1$, do $dB=-\frac{dC}{C^2}$. This gives
\begin{align}
\sqrt{2}\int_0^1 \frac{C}{4(\frac1C-1)^{1/4}}\frac{dC}{C^2}&=\sqrt{2}\int_0^1\frac{1}{4(1-C)^{1/4}C^{3/4}}dC\\
&=\frac{1}{2^{3/2}}B(\frac14,\frac34)
\end{align}
And if you look into it, $B(\frac14,\frac34)=\sqrt{2}\pi$. Therefore, the integral is equal to $\frac\pi2$ (which can be verified by numerical integration).
Note: the overall substitution works out to be $\frac{1-x^2}{x}=2\sqrt{\frac{1-C}C}$, or $x=\frac{1-\sqrt{1-C}}{\sqrt{C}}$.
