cos 2x + k
\(\frac{1}{2}\)cos 2x + k
\(-\frac{1}{2}\)cos 2x + k
-cos 2x + k
Correct answer is C
\(\int \sin 2x dx = \frac{1}{2} (-\cos 2x) + k\)
\(- \frac{1}{2} \cos 2x + k\)
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