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