Evaluate the given integral $\int {{e^x}(1 - \cot x + {{\cot }^2}x)dx} $=
${\text{A}}{\text{. }}{{\text{e}}^x}{\text{cot x + c}}$
${\text{B}}{\text{. - }}{{\text{e}}^x}{\text{cot x + c}}$
${\text{C}}{\text{. }}{{\text{e}}^x}{\text{cosec x + c}}$
${\text{D}}{\text{. - }}{{\text{e}}^x}{\text{cosec x + c}}$

Hint - Use the trigonometric formula, $1 + {\cot ^2}x = \cos e{c^2}x$ and then the integral will be in the form $\int {{e^x}[f(x) + f'(x)]dx = {e^x}f(x) + c} $.

Complete step-by-step solution -
We have been given the integral $\int {{e^x}(1 - \cot x + {{\cot }^2}x)dx} $.
Using the trigonometric formula, $1 + {\cot ^2}x = \cos e{c^2}x$, we get-
\[ \int {{e^x}(1 - \cot x + {{\cot }^2}x)dx} \\
  \Rightarrow \int {{e^x}(1 + {{\cot }^2}x - \cot x)dx} \\
  \Rightarrow \int {{e^x}( - \cot x + \cos e{c^2}x)dx} \\ \]
Now, $\dfrac{{d( - \cot x)}}{{dx}} = \cos e{c^2}x$ , which implies if $f(x) = - \cot x$ then $f'(x) = \cos e{c^2}x$
Therefore, we can say that the above integral is in the form $\int {{e^x}[f(x) + f'(x)]dx = {e^x}f(x) + c} $.
Therefore, the integral will be –
\[ \int {{e^x}( - \cot x + \cos e{c^2}x)dx} \\
   = - {e^x}\cot x + c \\ \]
Hence, the answer is option ${\text{B}}{\text{. - }}{{\text{e}}^x}{\text{cot x + c}}$.

Note – Whenever such types of questions appear, then first write the integral given to integrate. Then use the trigonometric formula on $1 +co{t^2}x $ which is equal to $\cos e{c^2}x$. Substituting this will make the integral into a simpler form, $\int {{e^x}[f(x) + f'(x)]dx = {e^x}f(x) + c} $. And then choose the correct option. Here students try to solve the given problem by multiplying the terms which is tough to solve in comparison to using the standard method of integration. So, there is a need to remember the standard formulas for solving these types of questions.