Evaluate \(\int\)(cos4x + sin3x)dx
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Correct Answer: Option C
Explanation:
\(\int (\cos 4x + \sin 3x) \mathrm d x\)
= \(\frac{1}{4} \sin 4x - \frac{1}{3} \cos 3x + k\)
\(\int (\cos 4x + \sin 3x) \mathrm d x\)
= \(\frac{1}{4} \sin 4x - \frac{1}{3} \cos 3x + k\)