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?
 A: We find 
\begin{align*}
\int_{0}^{\pi/2} \left(
\frac{1}{\sin^3{\theta}} - \frac{1}{\sin^2{\theta}}
\right)^{1/4} \cos{\theta}\, d\theta 
&= 2\int_0^{\pi/2} \sin^{-1/2}t \cos^{3/2}t \, dt,
    & \theta\rightarrow \arcsin \sin^2 t
\end{align*}
which is of the form 
$$2\int_0^{\pi/2} \sin^{2x-1}t \cos^{2y-1}t \, dt,$$
for $(x,y) = (1/4, 5/4)$.
A: Since you already received a good answer, another possible solution using the substitution proposed by @gt6989b in comments.
$$\int \left(\frac1{u^3} - \frac1{u^2}\right)^{1/4}\, du=2 \sqrt[4]{(1-u) u}+2 \sqrt[4]{u} \,\,\, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};u\right)$$
$$\int_0^1 \left(\frac1{u^3} - \frac1{u^2}\right)^{1/4}\, du=2\,_2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};1\right)=\frac{2 \sqrt{2 \pi }\, \Gamma \left(\frac{5}{4}\right)}{\Gamma
   \left(\frac{3}{4}\right)}=\frac{\left(\Gamma(1/4)\right)^2}{2\sqrt{\pi}}$$
A: Enforce the substitution $\sin(\theta)=\sin^2(t)$ to find that 
$$\begin{align} 
\int_0^{\pi/2} \left( \frac1{\sin^3(\theta)}-\frac1{\sin^2(\theta)}\right)^{1/4}\,\cos(\theta)\,d\theta&=2\int_0^{\pi/2}\cos^{3/2}(t)\sin^{-1/2}(t)\,dt\\\\
&=B(5/4,1/4)\\\\
&=\frac{\Gamma(5/4)\Gamma(1/4)}{\Gamma(3/2)}\\\\
&=\frac{\frac14 \Gamma^2(1/4)}{\frac12 \Gamma(1/2)}\\\\
&=\frac{\Gamma^2(1/4)}{2\sqrt{\pi}}
\end{align} $$
as was to be shown!
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
&\bbox[15px,#ffd]{\int_{0}^{\pi/2}\bracks{{1 \over \sin^{3}\pars{\theta}} -
{1 \over \sin^{2}\pars{\theta}}}^{1/4}\cos\pars{\theta}\,\dd\theta}
\\[5mm] \stackrel{t\ =\ \sin\pars{\theta}}{=}\,\,\,&
\int_{0}^{1}\pars{{1 \over t^{3}} - {1 \over t^{2}}}^{1/4}\,\dd t =
\int_{0}^{1}t^{-3/4}\pars{1 - t}^{1/4}\,\dd t =
{\Gamma\pars{1/4}\Gamma\pars{5/4} \over \Gamma\pars{3/2}}
\\[5mm] = &\
{\Gamma\pars{1/4}\bracks{\pars{1/4}\Gamma\pars{1/4}} \over
\Gamma\pars{1/2}/2} =
\bbox[15px,#ffd,border:1px solid navy]{{\Gamma^{2}\pars{1/4} \over 2\root{\pi}}}\ \approx\ 3.7081
\end{align}
Note that $\ds{\Gamma\pars{1/2} = \root{\pi}}$.
