# How can I show that the integral $\int_0^1 (1 - t^2)^{-3/4} dt$ is bounded?

Could someone please give me some hint on how can I show that the integral $\int_0^1 (1 - t^2)^{-3/4} dt$ is bounded? Thank you very much!

Note here that the integrand $(1-t^2)^{-3/4}$ can be factored as $(1-t)^{-3/4}(1+t)^{-3/4}$, the second part of which is bounded in the defining interval $(0,1)$.
Thus it suffices to show that the integral $\int_0^1(1-t)^{-3/4}dt$ is bounded
But then after linear change of variables $t \to 1-x$, $\int_0^1(1-t)^{-3/4}dt=\int_0^1x^{-3/4}dx$ is bounded as the primitive function of $x^{-3/4}$ is $4x^{1/4}$, bounded on $(0,1)$.
$$\int_{0}^{1}(1-t^2)^{-3/4}\,dt =\frac{1}{2}\int_{0}^{1}u^{-1/2}(1-u)^{-3/4}\,du = \frac{\Gamma\left(\frac{1}{2}\right)\,\Gamma\left(\frac{1}{4}\right)}{2\,\Gamma\left(\frac{3}{4}\right)}=\color{red}{\frac{\Gamma\left(\frac{1}{4}\right)^2}{2\sqrt{2\pi}}}.$$