2014
Mathematics
67ca00ce0c643c71d77dee1f
Evaluate ∫sin2xdx
A.
cos 2x + k
B.
1/2cos 2x + k
C.
−1/2 cos 2x + k
D.
-cos 2x + k
Correct Answer: −1/2 cos 2x + k
Explanation
∫sin2xdx =1/2(−cos2x) + k −1/2cos2x + k
∫sin2xdx =1/2(−cos2x) + k −1/2cos2x + kTHIS WEEK's
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