The value of int e^x left( frac{1 - sin x}{1 | vedclass

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(A) We need to evaluate the integral $I = \int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx$. Using trigonometric identities,$1 - \sin x = 1 -

Vedclass · Accepted Answer

$-e^x \cot \frac{x}{2} + c$,(where $c$ is a constant of integration)

Vedclass · Answer

(A) We need to evaluate the integral $I = \int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx$. Using trigonometric identities,$1 - \sin x = 1 - 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 - \cos x = 2 \sin^2 \frac{x}{2}$. Substituting these into the integral: $I = \int e^x \left( \frac{1 - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right) dx$ $I = \int e^x \left( \frac{1}{2 \sin^2 \frac{x}{2}} - \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right) dx$ $I = \int e^x \left( \frac{1}{2} \operatorname{cosec}^2 \frac{x}{2} - \cot \frac{x}{2} \right) dx$. Let $f(x) = -\cot \frac{x}{2}$. Then $f'(x) = -(-\operatorname{cosec}^2 \frac{x}{2}) \cdot \frac{1}{2} = \frac{1}{2} \operatorname{cosec}^2 \frac{x}{2}$. Since the integral is of the form $\int e^x (f(x) + f'(x)) dx = e^x f(x) + c$,$I = e^x (-\cot \frac{x}{2}) + c = -e^x \cot \frac{x}{2} + c$.

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

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