int left( frac{2 - sin 2x}{1 - cos 2x} right) | vedclass

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(A) Let $I = \int \left( \frac{2 - \sin 2x}{1 - \cos 2x} \right) e^x \, dx$. Using trigonometric identities $\sin 2x = 2 \sin x \cos x$ and $1 - \cos

Vedclass · Accepted Answer

$-e^x \cot x + c$

Vedclass · Answer

(A) Let $I = \int \left( \frac{2 - \sin 2x}{1 - \cos 2x} \right) e^x \, dx$. Using trigonometric identities $\sin 2x = 2 \sin x \cos x$ and $1 - \cos 2x = 2 \sin^2 x$,we get: $I = \int \left( \frac{2 - 2 \sin x \cos x}{2 \sin^2 x} \right) e^x \, dx$ $I = \int \left( \frac{1}{\sin^2 x} - \frac{\sin x \cos x}{\sin^2 x} \right) e^x \, dx$ $I = \int (\operatorname{cosec}^2 x - \cot x) e^x \, dx$ $I = \int e^x \operatorname{cosec}^2 x \, dx - \int e^x \cot x \, dx$. Applying integration by parts to the first integral $\int e^x \operatorname{cosec}^2 x \, dx$: Let $u = \cot x$,then $du = -\operatorname{cosec}^2 x \, dx$. $\int e^x \operatorname{cosec}^2 x \, dx = e^x(-\cot x) - \int e^x(-\cot x) \, dx = -e^x \cot x + \int e^x \cot x \, dx$. Substituting this back into the expression for $I$: $I = (-e^x \cot x + \int e^x \cot x \, dx) - \int e^x \cot x \, dx + c$ $I = -e^x \cot x + c$.

$\int \left( \frac{2 - \sin 2x}{1 - \cos 2x} \right) e^x \, dx$ is equal to

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