int e^{4x}(sin 3x - cos 3x) dx = | vedclass

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Question

(B) We use the standard integral formula: $\int e^{ax}(\sin bx - \cos bx) dx = \frac{e^{ax}}{a^2+b^2} [a \sin bx - b \cos bx - (a \cos bx + b \sin bx)

Vedclass · Accepted Answer

$\frac{e^{4x}}{25}(\sin 3x - 7 \cos 3x) + c$

Vedclass · Answer

(B) We use the standard integral formula: $\int e^{ax}(\sin bx - \cos bx) dx = \frac{e^{ax}}{a^2+b^2} [a \sin bx - b \cos bx - (a \cos bx + b \sin bx)] + c = \frac{e^{ax}}{a^2+b^2} [(a-b) \sin bx - (a+b) \cos bx] + c$. Here,$a = 4$ and $b = 3$. Substituting these values: $\int e^{4x}(\sin 3x - \cos 3x) dx = \frac{e^{4x}}{4^2+3^2} [(4-3) \sin 3x - (4+3) \cos 3x] + c$ $= \frac{e^{4x}}{16+9} [1 \sin 3x - 7 \cos 3x] + c$ $= \frac{e^{4x}}{25}(\sin 3x - 7 \cos 3x) + c$. Thus,the correct option is $B$.

$\int e^{4x}(\sin 3x - \cos 3x) dx = $

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