∫ e^-x(cot x + cosec^2 x)dx =

The correct answer is Option (2) → \(-e^{-x}\cot x + c\), where \(c\) is an arbitrary constant.

The integral is the sum of \(\int e^{-x}\cot x\,dx\) and \(\int e^{-x}\mathrm{cosec}^2 x\,dx\). By integration by parts on the second, with \(\int \mathrm{cosec}^2 x\,dx = -\cot x\) and \(\frac{d}{dx}e^{-x} = -e^{-x}\):

    \( \int e^{-x}\mathrm{cosec}^2 x\,dx = e^{-x}(-\cot x) - \int (-e^{-x})(-\cot x)\,dx \)

    \( = -e^{-x}\cot x - \int e^{-x}\cot x\,dx \)

Adding \(\int e^{-x}\cot x\,dx\) to both sides, the remaining integral cancels:

    \( \int e^{-x}(\cot x + \mathrm{cosec}^2 x)\,dx = -e^{-x}\cot x + c \)

Shortcut: Differentiate Option (2) by the product rule: \(\frac{d}{dx}(-e^{-x}\cot x) = e^{-x}\cot x + e^{-x}\mathrm{cosec}^2 x\), which is the integrand.