int e^x(1-cot x+cot^2 x) dx = | vedclass

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Question

(C) We know that $1 + \cot^2 x = \operatorname{cosec}^2 x$. Substituting this into the integral,we get: $\int e^x(1 - \cot x + \cot^2 x) dx = \int e^x

Vedclass · Accepted Answer

$-e^x \cdot \cot x + c$,where $c$ is a constant of integration.

Vedclass · Answer

(C) We know that $1 + \cot^2 x = \operatorname{cosec}^2 x$. Substituting this into the integral,we get: $\int e^x(1 - \cot x + \cot^2 x) dx = \int e^x(\operatorname{cosec}^2 x - \cot x) dx$. Let $f(x) = -\cot x$. Then $f'(x) = -(-\operatorname{cosec}^2 x) = \operatorname{cosec}^2 x$. Using the standard formula $\int e^x[f(x) + f'(x)] dx = e^x f(x) + c$,we have: $\int e^x(-\cot x + \operatorname{cosec}^2 x) dx = e^x(-\cot x) + c = -e^x \cdot \cot x + c$.

$\int e^x(1-\cot x+\cot^2 x) dx =$

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