$$log((sinx)^2)/tanx$$

$\sin x=t$とおけば$\cos xdx=dt$

よって

$$\begin{align*}\int \dfrac{\log(\sin^2x)}{\tan x}dx&=\int\log(\sin^2x)\cdot\dfrac{\cos x}{\sin x}dx\\&=\int\log t^2\cdot\dfrac{dt}{t}\\&=\dfrac{1}{2}\int\dfrac{2}{t}\log t^2dt\\&=\dfrac{1}{2}\int\left(\log t^2\right)'\log t^2dt\\&=\dfrac{1}{2}\cdot\dfrac{1}{2}\left(\log t^2\right)^2+C\\&=\dfrac{1}{4}\left\{\log (\sin^2x)\right\}^2+C\ \square\end{align*}$$

かと。