<s>[INST] Solve the integral using beta functions : $\int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2}\,dx$ I was solving questions on beta and gamma functions and then I came across this question.
$$ \int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2}\,dx $$
Generally in questions of beta functions integrals of the type $\int x^n(1−x^m) \, dx$ can be solved by substituting $x^m = t$, but in this case substitution is unconventional because if I put $1+x^4=t$, then I will not get the required form of beta function
I tried substituting $x^2 = \tan t$ and also tried substituting $x^2 = \cos t$ but on simplifying each of them further I could not reduce them to standard form of beta function.
 [/INST] Apply the substitution $x^4 = \frac{1-u}{1+u}$. Then
$$ I := \int_{0}^{1} \frac{(1-x^4)^{3/4}}{(1+x^4)^2} \, dx
= \frac{1}{2^{9/4}} \int_{0}^{1} u^{3/4}(1-u)^{-3/4} \, du
= \frac{1}{2^{9/4}} B(\tfrac{7}{4},\tfrac{1}{4})
= \frac{3\pi}{2^{15/4}}.$$

Explanations. First substitute $x^4 = t$ to obtain $ I = \frac{1}{4} \int_{0}^{1} \frac{(1-t)^{3/4}t^{-3/4}}{(1+t)^2} \, dx$. To transform this integral into the form of beta integral, we notice that 


*

*$-1$, $0$ and $1$ are the only branch points/poles of the integrand,

*$0$, $1$, $\infty$ are the only branch points/poles of the integrand of $\int_{0}^{1} u^{\alpha-1}(1-u)^{\beta-1} \, du$.


This leads to consider a Möbius transformation $t = f(u) = \frac{au+b}{cu+d}$ satisfying either 


*

*$f(\infty) = -1$, $f(0) = 0$ and $f(1) = 1$ and $ad-bc > 0$, or

*$f(\infty) = -1$, $f(0) = 1$ and $f(1) = 0$ and $ad-bc < 0$. 


The first condition yields  $f(u) = \frac{u}{2-u}$ and the second condition yields $f(u) = \frac{1-u}{1+u}$. Both can be used, since one condition is simply transformed to the other by $z \mapsto 1-z$.
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