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
$$ \begin{eqnarray} \frac{ \partial }{ \partial n }B(n,m) &=& 2\int_{0}^{\frac{\pi}{2}}\frac{ \partial }{ \partial n }\sin^{2n-1}x\cos^{2m-1}xdx\\ &=& 4\int_{0}^{\frac{\pi}{2}}\sin^{2n-1}x\cos^{2m-1}x\ln(\sin x) dx \end{eqnarray} $$
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
$$ \begin{eqnarray} \int_{0}^{\frac{\pi}{2}}\frac{\ln{\sin{x}}}{\sqrt{\cos{x}}}dx &=& \frac{1}{4}B\left(\frac{1}{2},\frac{1}{4}\right) \int_{0}^{1} \frac{t^{-\frac{1}{4}}-t^{-\frac{1}{2}}}{1-t}dt\\ &=& \frac{1}{4}\int_{0}^{1}x^{-\frac{1}{2}}(1-x)^{-\frac{3}{4}}dx \cdot 4\int_{0}^{1} \frac{y^2-y}{1-y^4}dy \quad (y=t^{\frac{1}{4}})\\ &=& \frac{1}{4}\cdot 4\int_{0}^{1}\frac{1}{\sqrt{1-u^4}}du \cdot 2\int_{0}^{1} \frac{-2y}{(y+1)(y^2+1)}dy\quad (u=(1-x)^{\frac{1}{4}})\\ &=& \frac{\varpi}{2}\cdot 2\int_{0}^{1}\frac{1}{y+1}-\frac{y+1}{y^2+1}dy\\ &=& \frac{\varpi}{2} \left( 2\int_{0}^{1}\frac{dy}{y+1}-2\int_{0}^{1}\frac{y}{y^2+1}dy-2\int_{0}^{1}\frac{1}{y^2+1}dy \right)\\ &=& \frac{\varpi}{2} \left( 2\log2-\log2-\frac{\pi}{2} \right)\\ &=& \frac{\varpi}{2} \left(\log2-\frac{\pi}{2} \right)\\ \end{eqnarray} $$