Question

$\int e^x\left(\cot x+\log \sin x\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int e^x\left(\cot x+\log \sin x\right)dx \\ \mathrm{Here},f\left(x\right)=\log \sin x\mathrm{Put}{e}^{x}f\left(x\right)=t \\ \Rightarrow f\text{'}\left(x\right)=\cot x \\ \mathrm{let}{e}^x\log \sin x=t \\ \mathrm{Diff}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}x \\ {e}^x\log \left(\sin x\right)+e^x\times \frac{1}{\sin x}\times \cos x=\frac{dt}{dx} \\ \Rightarrow \left[e^x\log \left(\sin x\right)+e^x\cot x\right]dx=dt \\ \Rightarrow e^x\left(\cot x+\log \sin x\right)dx=dt \\ \therefore \int e^x\left(\cot x+\log \sin x\right)dx=\int dt \\ =t+C \\ =e^x\log \sin x+C \end{gathered}$$