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$\int\left(\frac{2-\sin 2 x}{1-\cos 2 x}\right) e^x d x$ is equal to
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The correct answer is:
$-e^x \cot x+c$
Let $\quad \begin{aligned} I & =\int\left(\frac{2-\sin 2 x}{1-\cos 2 x}\right) e^x d x \\ & =\int\left(\frac{2-2 \sin x \cos x}{2 \sin ^2 x}\right) e^x d x \\ & =\int \operatorname{cosec}^2 x e^x d x-\int \cot x e^x d x \\ & =-\cot x e^x-\int(-\cot x) e^x d x \\ & \quad-\int \cot x e^x d x+c \\ & =-\cot x e^x+c\end{aligned}$
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