int {{e^x}left( {frac{{1 - sin x}}{{1 - cos x | vedclass

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(C) We know that $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$. Given integral is $I = \int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx$. Using

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

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(C) We know that $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$. Given integral is $I = \int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx$. Using trigonometric identities $1 - \sin x = 1 - 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 - \cos x = 2 \sin^2 \frac{x}{2}$,we get: $I = \int e^x \left( \frac{1 - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right) dx$ $I = \int e^x \left( \frac{1}{2 \sin^2 \frac{x}{2}} - \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right) dx$ $I = \int e^x \left( \frac{1}{2} \csc^2 \frac{x}{2} - \cot \frac{x}{2} \right) dx$ Let $f(x) = - \cot \frac{x}{2}$. Then $f'(x) = - \left( - \csc^2 \frac{x}{2} \cdot \frac{1}{2} \right) = \frac{1}{2} \csc^2 \frac{x}{2}$. Thus,$I = \int e^x [f(x) + f'(x)] dx = e^x f(x) + C = -e^x \cot \frac{x}{2} + C$.

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

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