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

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(C) We are given the integral $I = \int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx$. Using trigonometric identities $\sin x = 2 \sin \frac{x}

Vedclass · Accepted Answer

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

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(C) We are given the integral $I = \int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx$. Using trigonometric identities $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 - \cos x = 2 \sin^2 \frac{x}{2}$, we get: $I = \int e^x \left( \frac{1 - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right) dx$ $I = \int e^x \left( \frac{1}{2 \sin^2 \frac{x}{2}} - \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right) dx$ $I = \int e^x \left( \frac{1}{2} \operatorname{cosec}^2 \frac{x}{2} - \cot \frac{x}{2} \right) dx$ Let $g(x) = -\cot \frac{x}{2}$. Then $g'(x) = -(-\operatorname{cosec}^2 \frac{x}{2} \cdot \frac{1}{2}) = \frac{1}{2} \operatorname{cosec}^2 \frac{x}{2}$. Since the integral is of the form $\int e^x (g(x) + g'(x)) dx = e^x g(x) + c$, we have: $I = e^x \left( -\cot \frac{x}{2} \right) + c = -e^x \cot \frac{x}{2} + c$. Thus, $f(x) = -e^x \cot \left( \frac{x}{2} \right)$.

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

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