\(\int e^{2x}\cos 3x\,dx\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{e^{2x}}{13}\left(2\cos 3x+3\sin 3x\right)+C\)
Let \(I = \int e^{2x}\cos 3x\,dx\). Using integration by parts with \(\cos 3x\) as the first function and \(e^{2x}\) as the second:
\( \displaystyle I = \frac{1}{2}e^{2x}\cos 3x+\frac{3}{2}\int e^{2x}\sin 3x\,dx \)
By parts again on the last integral:
\( \displaystyle \int e^{2x}\sin 3x\,dx = \frac{1}{2}e^{2x}\sin 3x-\frac{3}{2}I \)
So the integral returns:
\( \displaystyle I = \frac{1}{2}e^{2x}\cos 3x+\frac{3}{4}e^{2x}\sin 3x-\frac{9}{4}I \)
\( \displaystyle \frac{13}{4}I = \frac{e^{2x}}{4}\left(2\cos 3x+3\sin 3x\right) \)
\( \displaystyle I = \frac{e^{2x}}{13}\left(2\cos 3x+3\sin 3x\right)+C \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{e^{2x}}{13}\left(2\cos 3x-3\sin 3x\right)+C\) — Incorrect. Its derivative is \(-\frac{e^{2x}}{13}\left(5\cos 3x+12\sin 3x\right)\).
Option (2) \(\displaystyle\frac{e^{2x}}{13}\left(3\cos 3x+2\sin 3x\right)+C\) — Incorrect. The coefficients 2 and 3 are interchanged; its derivative is \(\frac{e^{2x}}{13}\left(12\cos 3x-5\sin 3x\right)\).
Option (3) \(\displaystyle\frac{e^{2x}}{5}\left(2\cos 3x+3\sin 3x\right)+C\) — Incorrect. The divisor is \(2^2+3^2 = 13\), not 5.