If ∫ (cot x - cosec² x)eˣ dx = eˣ f(x) + c then f(x) will be:

If $\int (\cot x - \operatorname{cosec}^2 x)e^x dx = e^x f(x) + c$ then $f(x)$ will be:

(a)$\cot x + \operatorname{cosec} x$
(b)$\cot^2 x$
(c)$\cot x$
(d)$\operatorname{cosec} x$

Answer

Correct option: (c)

Answer: (c) $\cot x$ Using $\int e^x\left[f(x) + f'(x)\right]dx = e^x f(x) + C$ with $f(x) = \cot x$, $f'(x) = -\operatorname{cosec}^2 x$. So $\int(\cot x - \operatorname{cosec}^2 x)e^x dx = e^x\cot x + C$, hence $f(x) = \cot x$.