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$\int e^{- x} \cos e c x (\cot x + 1) d x =$

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

$e^{x} \cos e c x + c$

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

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

answer is D.

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Detailed Solution

$\int e^{- x} \cos e c x (\cot x + 1) d x$

$= \int e^{- x} (\cos e c x . \cot x + \cos e c x) d x$

$= - \int e^{- x} [(- 1) \cos e c x + (- \cos e c x . c o t x)] d x$

$= - e^{- x} [\cos e c x] + c$

$(∵ \int e^{g (x)} [g' (x) f (x) + f' (x)] d x = e^{g (x)} f (x) + c)$

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