AIが出力した積分を解く

$$B(n,m)={\int }_0^1{x}^{n-1}(1-x{)}^{m-1}dx=2{\int }_0^{\frac{\pi }{2}}{\sin }^{2n-1}x{\cos }^{2m-1}xdx$$
より
$$\begin{array}{rcl}\frac{\partial }{\partial n}B(n,m)& =& 2{\int }_0^{\frac{\pi }{2}}\frac{\partial }{\partial n}{\sin }^{2n-1}x{\cos }^{2m-1}xdx\\ & =& 4{\int }_0^{\frac{\pi }{2}}{\sin }^{2n-1}x{\cos }^{2m-1}x\ln (\sin x)dx\end{array}$$
公式$1$公式$2$と合わせて
$${\int }_0^{\frac{\pi }{2}}{\sin }^{2n-1}x{\cos }^{2m-1}x\ln (\sin x)dx=\frac{1}{4}B(n,m)(\psi (n)-\psi (n+m))=\frac{1}{4}B(n,m){\int }_0^1\frac{t^{n+m-1}-t^{n-1}}{1-t}dt$$
これに$n=\frac{1}{2},m=\frac{1}{4}$を代入すれば求めたい積分を得られる
$$\begin{array}{rcl}{\int }_0^{\frac{\pi }{2}}\frac{\ln \sin x}{\sqrt{\cos x}}dx& =& \frac{1}{4}B(\frac{1}{2},\frac{1}{4}){\int }_0^1\frac{t^{-\frac{1}{4}}-t^{-\frac{1}{2}}}{1-t}dt\\ & =& \frac{1}{4}{\int }_0^1{x}^{-\frac{1}{2}}(1-x{)}^{-\frac{3}{4}}dx\cdot 4{\int }_0^1\frac{y^2-y}{1-y^4}dy(y=t^{\frac{1}{4}})\\ & =& \frac{1}{4}\cdot 4{\int }_0^1\frac{1}{\sqrt{1-u^4}}du\cdot 2{\int }_0^1\frac{-2y}{(y+1)(y^2+1)}dy(u=(1-x{)}^{\frac{1}{4}})\\ & =& \frac{\varpi }{2}\cdot 2{\int }_0^1\frac{1}{y+1}-\frac{y+1}{y^2+1}dy\\ & =& \frac{\varpi }{2}(2{\int }_0^1\frac{dy}{y+1}-2{\int }_0^1\frac{y}{y^2+1}dy-2{\int }_0^1\frac{1}{y^2+1}dy)\\ & =& \frac{\varpi }{2}(2\log 2-\log 2-\frac{\pi }{2})\\ & =& \frac{\varpi }{2}(\log 2-\frac{\pi }{2})\end{array}$$