Question

$\int e^x\left(\cot x-{\mathrm{cosec}}^{2}x\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int e^x\left(\cot x-{\mathrm{cosec}}^{2}x\right)dx \\ \mathrm{here}f\left(x\right)=\cot x\mathrm{put}{e}^{x}f\left(x\right)=t \\ f\text{'}\left(x\right)=-{\mathrm{cosec}}^{2}x \\ \mathrm{let}{e}^x\cot x=t \\ \mathrm{Diff}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}x \\ {e}^x\cot x+e^x\left(-{\mathrm{cosec}}^{2}x\right)=\frac{dt}{dx} \\ \Rightarrow e^x\left(\cot x-{\mathrm{cosec}}^{2}x\right)dx=dt \\ \therefore \int e^x\left(\cot x-{\mathrm{cosec}}^{2}x\right)dx=\int dt \\ =t+C \\ =e^x\cot x+C \end{gathered}$$