Question

$\int \frac{\cot \mathrm{x}}{\log \sin x}dx$

Answer 1

$$\begin{gathered} \mathrm{Note}:\mathrm{Here}\mathrm{we}\mathrm{are}\mathrm{considering}\log x\mathrm{as}{\log }_{\mathrm{e}}x \\ \mathrm{Let}I=\int \frac{\cot x}{\log \sin x}dx \\ \mathrm{Putting}\log \sin x=t \\ \Rightarrow \cot x=\frac{dt}{dx} \\ \Rightarrow \cot xdx=dt \\ \therefore I=\int \frac{1}{t}dt \\ =\log \left|t\right|+C \\ =\log \left|\log \sin x\right|+C\left[\because t=\log \sin x\right] \end{gathered}$$