sin4x - cos3x + k
sin4x + cos3x + k
\(\frac{1}{4}\)sin4x - \(\frac{1}{3}\)cos3x + k
\(\frac{1}{4}\)sin4x + \(\frac{1}{3}\)cos3x + k
Correct answer is C
\(\int (\cos 4x + \sin 3x) \mathrm d x\)
= \(\frac{1}{4} \sin 4x - \frac{1}{3} \cos 3x + k\)
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