int {{e^x}(1 - cot x + {{cot }^2}x),dx} equa | vedclass

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

Vedclass · Accepted Answer

$-{e^x}\cot x + c$

Vedclass · Answer

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

$\int {{e^x}(1 - \cot x + {{\cot }^2}x)\,dx} $ equals

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