int e^{-2x} sin 3x , dx = | vedclass

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(D) Let $I = \int e^{-2x} \sin 3x \, dx$. Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = \sin 3x$ and $dv = e^{-2x} \, dx$. T

Vedclass · Accepted Answer

$-\frac{1}{13} e^{-2x} [2 \sin 3x + 3 \cos 3x] + c$

Vedclass · Answer

(D) Let $I = \int e^{-2x} \sin 3x \, dx$. Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = \sin 3x$ and $dv = e^{-2x} \, dx$. Then $du = 3 \cos 3x \, dx$ and $v = -\frac{1}{2} e^{-2x}$. $I = -\frac{1}{2} e^{-2x} \sin 3x - \int (-\frac{1}{2} e^{-2x}) (3 \cos 3x) \, dx$ $I = -\frac{1}{2} e^{-2x} \sin 3x + \frac{3}{2} \int e^{-2x} \cos 3x \, dx$. Now,integrate $\int e^{-2x} \cos 3x \, dx$ by parts again: Let $u = \cos 3x$ and $dv = e^{-2x} \, dx$. Then $du = -3 \sin 3x \, dx$ and $v = -\frac{1}{2} e^{-2x}$. $\int e^{-2x} \cos 3x \, dx = -\frac{1}{2} e^{-2x} \cos 3x - \int (-\frac{1}{2} e^{-2x}) (-3 \sin 3x) \, dx$ $= -\frac{1}{2} e^{-2x} \cos 3x - \frac{3}{2} \int e^{-2x} \sin 3x \, dx$. Substituting this back into the expression for $I$: $I = -\frac{1}{2} e^{-2x} \sin 3x + \frac{3}{2} [-\frac{1}{2} e^{-2x} \cos 3x - \frac{3}{2} I]$ $I = -\frac{1}{2} e^{-2x} \sin 3x - \frac{3}{4} e^{-2x} \cos 3x - \frac{9}{4} I$ $I + \frac{9}{4} I = -\frac{1}{4} e^{-2x} [2 \sin 3x + 3 \cos 3x]$ $\frac{13}{4} I = -\frac{1}{4} e^{-2x} [2 \sin 3x + 3 \cos 3x]$ $I = -\frac{1}{13} e^{-2x} [2 \sin 3x + 3 \cos 3x] + c$.

$\int e^{-2x} \sin 3x \, dx = $

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