Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int e^x\left(\frac{1+\sin x}{1+\cos x}\right)dx \\ =\int e^x\left(\frac{1}{1+\cos x}+\frac{\sin x}{1+\cos x}\right)dx \\ =\int e^x\left(\frac{1}{2{\cos }^2{\displaystyle \frac{x}{2}}}+\frac{2\sin {\displaystyle \frac{x}{2}}\cos {\displaystyle \frac{x}{2}}}{2{\cos }^2{\displaystyle \frac{x}{2}}}\right)dx \\ =\int e^x\left(\frac{1}{2}{\sec }^2\frac{x}{2}+\tan \frac{x}{2}\right)dx \\ \mathrm{Putting}{e}^x\tan \frac{x}{2}=t \\ \mathrm{Diff}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.x \\ {e}^x·\tan \left(\frac{x}{2}\right)+e^x\times \frac{1}{2}{\sec }^2\frac{x}{2}=\frac{dt}{dx} \\ \Rightarrow e^x\left[\tan \frac{x}{2}+\frac{1}{2}{\sec }^2\left(\frac{x}{2}\right)\right]dx=dt \\ \therefore \int e^x\left(\frac{1}{2}{\sec }^2\frac{x}{2}+\tan \frac{x}{2}\right)dx=\int dt \\ =t+C \\ =e^x\tan \left(\frac{x}{2}\right)+C \end{gathered}$$