<s>[INST] Show that $\int_{0}^{\pi/2} (\frac{1}{\sin^3{\theta}} - \frac{1}{\sin^2{\theta}})^{1/4} \cos{\theta} d\theta = \frac{(\Gamma(1/4))^2}{2\sqrt{\pi}}$ Well, I have shown that $B(n, n+1) = \frac{(\Gamma(n))^2}{2\sqrt{2n}}$
From there I could deduce that $B(1/4, 5/4) = \frac{(\Gamma(1/4))^2}{2\sqrt{\pi}}$, then $n=1/4$.
I also know that $B(x, y) = \frac{(\Gamma(x))(\Gamma(y))} {\Gamma(x+y)} = 2 \int_{0}^{\pi/2} \sin^{2x-1}{\theta} \cos^{2y-1}{\theta} d\theta$. 
So I suppose I should reduce the given integral to a form similar to the one above. Is there any trigonometric property that can help me that or am I seeing it wrong?
 [/INST] Since you already received a good answer, another possible solution using the substitution proposed by @gt6989b in comments.
$$\int \left(\frac1{u^3} - \frac1{u^2}\right)^{1/4}\, du=2 \sqrt[4]{(1-u) u}+2 \sqrt[4]{u} \,\,\, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};u\right)$$
$$\int_0^1 \left(\frac1{u^3} - \frac1{u^2}\right)^{1/4}\, du=2\,_2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};1\right)=\frac{2 \sqrt{2 \pi }\, \Gamma \left(\frac{5}{4}\right)}{\Gamma
   \left(\frac{3}{4}\right)}=\frac{\left(\Gamma(1/4)\right)^2}{2\sqrt{\pi}}$$
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