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A.
cos 2x + k
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B.
\(\frac{1}{2}\)cos 2x + k
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C.
\(-\frac{1}{2}\)cos 2x + k
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D.
-cos 2x + k
Correct Answer: Option C
Explanation
\(\int \sin 2x dx = \frac{1}{2} (-\cos 2x) + k\)
\(- \frac{1}{2} \cos 2x + k\)
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