If $\int e^x\left(\frac{1-\sin x}{1-\cos x}\right) d x=f(x)+$ Constant, then $f(x)$ is equal to
Answer & explanation
Correct answer: option 3
We have,
$I=\int e^x\left(\frac{1-\sin x}{1-\cos x}\right) d x$
$\Rightarrow I=\int e^x\left(\frac{1}{2} cosec^2 \frac{x}{2}-\cot \frac{x}{2}\right) d x$
$\Rightarrow I=\int e^x\left(-\cot \frac{x}{2}+\frac{1}{2} cosec^2 \frac{x}{2}\right) d x=-e^x \cot \frac{x}{2}+C$