Question

$\int \frac{1+\cot x}{x+\log \sin x}dx$

Answer 1

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