int e^{3x} sin(4x-5) dx = . . . . . . + C | vedclass

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Question

(C) We use the standard formula: $\int e^{ax} \sin(bx+c) dx = \frac{e^{ax}}{a^2+b^2} [a \sin(bx+c) - b \cos(bx+c)] + C$. Here,$a = 3$,$b = 4$,and the

Vedclass · Accepted Answer

$\frac{e^{3x}}{25}[3 \sin(4x-5) - 4 \cos(4x-5)]$

Vedclass · Answer

(C) We use the standard formula: $\int e^{ax} \sin(bx+c) dx = \frac{e^{ax}}{a^2+b^2} [a \sin(bx+c) - b \cos(bx+c)] + C$. Here,$a = 3$,$b = 4$,and the constant term is $-5$. Substituting these values into the formula: $\int e^{3x} \sin(4x-5) dx = \frac{e^{3x}}{3^2 + 4^2} [3 \sin(4x-5) - 4 \cos(4x-5)] + C$. $= \frac{e^{3x}}{9 + 16} [3 \sin(4x-5) - 4 \cos(4x-5)] + C$. $= \frac{e^{3x}}{25} [3 \sin(4x-5) - 4 \cos(4x-5)] + C$. Thus,the correct option is $C$.

$\int e^{3x} \sin(4x-5) dx = $ . . . . . . $+ C$

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