Find ∫ e^x ≤ft( (1-sin x)/(1-cos x) ) dx

The correct answer is Option (2) → $-e^x \cot \left( \frac{x}{2} \right) + C$

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

$I = \int e^x \left( \frac{1 - 2\sin \frac{x}{2} \cos \frac{x}{2}}{2\sin^2 \frac{x}{2}} \right) dx$

$I = \int e^x \left( \frac{1}{2\sin^2 \frac{x}{2}} - \frac{2\sin \frac{x}{2} \cos \frac{x}{2}}{2\sin^2 \frac{x}{2}} \right) dx$

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

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

Let $f(x) = -\cot \frac{x}{2}$

$f'(x) = +\frac{1}{2} \text{cosec}^2 \frac{x}{2}$

Using $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$:

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

$= -e^x \cot \frac{x}{2} + C$