Integrate the function, $\int{{{e}^{-x}}\csc x\left( 1+\cot x \right)dx}$
(a) ${{e}^{-x}}\csc x+C$
(b) $-{{e}^{-x}}\csc x+C$
(c) $-{{e}^{-x}}\left( \csc x+\cot x \right)+C$
(d) $-{{e}^{-x}}\left( \csc x+tanx \right)+C$
(e) $-{{e}^{-x}}\sec x+C$

Hint: First convert ${{e}^{-x}}\csc x\left( 1+\cot x \right)$ to $-{{e}^{-x}}\left(- \left( \csc x\cot x \right)-\csc x \right)$ then apply the formula,
$\int{{{e}^{-x}}\left( f'\left( x \right)-f\left( x \right) \right)={{e}^{-x}}f\left( x \right)+C}$, here f(x) is a function and ${{f}^{'}}\left( x \right)$ is obtained on differentiating.

Complete Step-by-Step solution:
In these types of question we have to use identity,
$\int{{{e}^{-x}}\left( f'\left( x \right)-f\left( x \right) \right)dx={{e}^{-x}}f\left( x \right)}$
Here in this case if f(x) be a function of x and $f'\left( x \right)$ will be a function we got after differentiation of f(x) with respect to x and it is of form ${{e}^{-x}}\left( {{f}^{'}}\left( x \right)-f\left( x \right) \right)$ then after integration it will become ${{e}^{-x}}f\left( x \right)$ .
In the given question, we have ${{e}^{-x}}\csc x\left( 1+\cot x \right)$ then we can write it as,
$-{{e}^{-x}}\left( \left( \csc x\cot x \right)-\csc x \right)........(i)$
Let us consider $\csc x$ be f(x). Then on differentiating f(x) we get,
$f'\left( x \right)=-\csc x\cot x$
So, we can write the expression (i) as,
$-{{e}^{-x}}\left( {{f}^{'}}\left( x \right)-f\left( x \right) \right)$
Here $f\left( x \right)=\csc x$.
So, applying the formula, $\int{{{e}^{-x}}\left( f'\left( x \right)-f\left( x \right) \right)dx={{e}^{-x}}f\left( x \right)}$, the integration of $-{{e}^{-x}}\left( \left( \csc x\cot x \right)-\csc x \right)$ will be represented as,
\[\int{-{{e}^{-x}}\left( \left( -\csc x\cot x \right)-\csc x \right)=}-{{e}^{-x}}\left( \csc x \right)+C\]
Therefore the value of the integration of function ${{e}^{-x}}\csc x\left( 1+\cot x \right)$ is $-{{e}^{-x}}\left( cscx \right)+C$.
Hence, the correct answer is option (b).

Note: Student should be careful while transforming the expression ${{e}^{-x}}\csc x\left( 1+\cot x \right)$ to \[-{{e}^{-x}}\left( \left( -\csc x\cot x \right)-\csc x \right)\]. This is the area where students generally can’t get the idea what to do, they should visualize using practise.