If int e^{3x} cos 4x ,dx = e^{3x} (A sin 4x + | vedclass

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

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$3A = 4B$

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(C) We use the standard integration formula: $\int e^{ax} \cos(bx) \,dx = \frac{e^{ax}}{a^2 + b^2} (a \cos(bx) + b \sin(bx)) + c$. Here,$a = 3$ and $b = 4$. Substituting these values into the formula: $\int e^{3x} \cos 4x \,dx = \frac{e^{3x}}{3^2 + 4^2} (3 \cos 4x + 4 \sin 4x) + c$ $= \frac{e^{3x}}{25} (4 \sin 4x + 3 \cos 4x) + c$ $= e^{3x} (\frac{4}{25} \sin 4x + \frac{3}{25} \cos 4x) + c$. Comparing this with the given expression $e^{3x} (A \sin 4x + B \cos 4x) + c$,we get: $A = \frac{4}{25}$ and $B = \frac{3}{25}$. Now,checking the options: $3A = 3(\frac{4}{25}) = \frac{12}{25}$ and $4B = 4(\frac{3}{25}) = \frac{12}{25}$. Thus,$3A = 4B$ is correct.

If $\int e^{3x} \cos 4x \,dx = e^{3x} (A \sin 4x + B \cos 4x) + c$,then:

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