∫sin2x dx = (a) k + 2cos2x(b) cos2x/2 + k(c) k - cos2x/2(d) k - cos – InterviewSolution

1.	∫sin2x dx = (a) k + 2cos2x(b) cos2x/2 + k(c) k - cos2x/2(d) k - cos2x/3
Answer» Answer is (b) cos2x/2 + k

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