Question

$\int e^x\left(\frac{\sin 4x-4}{1-\cos 4x}\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int e^x\left[\frac{\sin 4x-4}{1-\cos 4x}\right]dx \\ =\int e^x\left[\frac{2\sin 2x\cos 2x}{2{\sin }^2\left(2x\right)}-\frac{4}{2{\sin }^{2}2x}\right]dx \\ =\int e^x\left[\cot \left(2x\right)-2{\mathrm{cosec}}^2\left(2x\right)\right]dx \\ \mathrm{Here},f\left(x\right)=\cot \left(2x\right) \\ \Rightarrow f\text{'}\left(x\right)=-2{\mathrm{cosec}}^2\left(2x\right) \\ \mathrm{Put}{e}^{x}f\left(x\right)=t \\ \mathrm{let}{e}^x\cot \left(2x\right)=t \\ \mathrm{Diff}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}x \\ {e}^x\cot \left(2x\right)+e^x\times \left[-2{\mathrm{cosec}}^2\left(2x\right)\right]=\frac{dt}{dx} \\ \Rightarrow e^x\left[\cot \left(2x\right)-2{\mathrm{cosec}}^2\left(2x\right)\right]dx=dt \\ \therefore \int e^x\left[\cot 2x-2{\mathrm{cosec}}^{2}2x\right]dx=\int dt \\ \Rightarrow I=t+C \\ =e^x\cot \left(2x\right)+C \end{gathered}$$