\(\int\left(\cos 3x+6e^{2x}\right)dx\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{3}\sin 3x+3e^{2x}+C\)
Using the standard integrals \(\int\cos ax\,dx = \frac{1}{a}\sin ax\) and \(\int e^{ax}\,dx = \frac{1}{a}e^{ax}\):
\( \displaystyle \int\cos 3x\,dx = \frac{1}{3}\sin 3x \)
\( \displaystyle \int 6e^{2x}\,dx = 6\times\frac{1}{2}e^{2x} = 3e^{2x} \)
Adding:
\( \displaystyle \int\left(\cos 3x+6e^{2x}\right)dx = \frac{1}{3}\sin 3x+3e^{2x}+C \)
Why the other options are incorrect
Option (2) \(\displaystyle -\frac{1}{3}\sin 3x+3e^{2x}+C\) — Incorrect. The derivative of \(\sin 3x\) is \(+3\cos 3x\), so no minus sign appears; the minus belongs to \(\int\sin ax\,dx\).
Option (3) \(3\sin 3x+12e^{2x}+C\) — Incorrect. It multiplies by a instead of dividing by a; its derivative is \(9\cos 3x+24e^{2x}\).
Option (4) \(\displaystyle\frac{1}{3}\sin 3x+12e^{2x}+C\) — Incorrect. The exponential term should be divided by 2, not multiplied; its derivative is \(24e^{2x}\).