Question

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

(a) $-e^x\tan \frac{x}{2}+C$




(c) $-\frac{1}{2}{e}^x\tan \frac{x}{2}+C$

(d) $-\frac{1}{2}{e}^x\cot \frac{x}{2}+C$

Answer 1

(b) $-e^x\cot \frac{x}{2}+C$

$$\begin{gathered} \mathrm{Let}I=\int e^x\left(\frac{1-\sin x}{1-\cos x}\right)dx \\ \Rightarrow \int e^x\left(\frac{1}{1-\cos x}-\frac{\sin x}{1-\cos x}\right)dx \\ \Rightarrow \int e^x\left(\frac{1}{2{\sin }^2{\displaystyle \frac{x}{2}}}-\frac{2\sin {\displaystyle \frac{x}{2}}\cos {\displaystyle \frac{x}{2}}}{2{\sin }^2{\displaystyle \frac{x}{2}}}\right)dx \\ \Rightarrow \int e^x\left(\frac{1}{2}{\mathrm{cosec}}^2\frac{x}{2}-\cot \frac{x}{2}\right)dx \\ \mathrm{As},\mathrm{we}\mathrm{know}\mathrm{that}\int e^x\left\{f\left(x\right)+f\text{'}\left(x\right)\right\}dx=e^{x}f\left(x\right)+C \\ \therefore I=-e^x\cot \left(\frac{x}{2}\right)+C \end{gathered}$$