If $\int e^x\left(\frac{1-\sin x}{1-\cos x}\right) d x=f(x)+$ Constant, then $f(x)$ is equal to

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

We have,

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

$\Rightarrow I=\int e^x\left(\frac{1}{2} cosec^2 \frac{x}{2}-\cot \frac{x}{2}\right) d x$

$\Rightarrow I=\int e^x\left(-\cot \frac{x}{2}+\frac{1}{2} cosec^2 \frac{x}{2}\right) d x=-e^x \cot \frac{x}{2}+C$