If I = int e^x sin 2x , dx,then for what valu | vedclass

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(C) Let $I = \int e^x \sin 2x \, dx$. Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = \sin 2x$ and $dv = e^x \, dx$. Then $du

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$5$

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(C) Let $I = \int e^x \sin 2x \, dx$. Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = \sin 2x$ and $dv = e^x \, dx$. Then $du = 2 \cos 2x \, dx$ and $v = e^x$.$I = e^x \sin 2x - \int e^x (2 \cos 2x) \, dx = e^x \sin 2x - 2 \int e^x \cos 2x \, dx$.Now,apply integration by parts again to $\int e^x \cos 2x \, dx$. Let $u = \cos 2x$ and $dv = e^x \, dx$. Then $du = -2 \sin 2x \, dx$ and $v = e^x$.$\int e^x \cos 2x \, dx = e^x \cos 2x - \int e^x (-2 \sin 2x) \, dx = e^x \cos 2x + 2 \int e^x \sin 2x \, dx$.Substituting this back into the expression for $I$:$I = e^x \sin 2x - 2(e^x \cos 2x + 2I) = e^x \sin 2x - 2e^x \cos 2x - 4I$.$I + 4I = e^x(\sin 2x - 2 \cos 2x) + C$.$5I = e^x(\sin 2x - 2 \cos 2x) + C$.Comparing this with $KI = e^x(\sin 2x - 2 \cos 2x) + C$,we get $K = 5$.

If $I = \int e^x \sin 2x \, dx$,then for what value of $K$ is $KI = e^x(\sin 2x - 2\cos 2x) + C$?

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