Step 1: Guess the antiderivative from the structure.
An $e^{\cot x}$ multiplying a sum suggests $F(x)=-2e^{\cot x}\log(\operatorname{cosec} x)$; we verify by differentiating.
Step 2: Differentiate the exponential factor.
Since $\dfrac{d}{dx}\cot x=-\operatorname{cosec}^2 x$, we have $\dfrac{d}{dx}e^{\cot x}=-\operatorname{cosec}^2 x\,e^{\cot x}$.
Step 3: Differentiate the log factor.
$\dfrac{d}{dx}\log(\operatorname{cosec} x)=\dfrac{1}{\operatorname{cosec} x}\cdot(-\operatorname{cosec} x\cot x)=-\cot x$.
Step 4: Apply the product rule.
$F'(x)=-2\big[(-\operatorname{cosec}^2 x\,e^{\cot x})\log(\operatorname{cosec} x)+e^{\cot x}(-\cot x)\big]=2e^{\cot x}\operatorname{cosec}^2 x\log(\operatorname{cosec} x)+2e^{\cot x}\cot x$.
Step 5: Rewrite in the integrand's shape.
Since $\operatorname{cosec}^2 x=\dfrac{1}{\sin^2 x}$ and $2\cot x=\dfrac{2\cos x}{\sin x}=\dfrac{\sin 2x}{\sin^2 x}$, we get $F'(x)=\dfrac{e^{\cot x}}{\sin^2 x}\big(2\log\operatorname{cosec} x+\sin 2x\big)$.
Step 6: Conclude.
This is exactly the integrand, so the integral is $F(x)+C$.
\[ \boxed{-2e^{\cot x}\log(\operatorname{cosec} x)+C} \]