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Question

# If the reduction formula for In=∫sin nxcos xdx is given by In+In−2=−2cos{(n−1)x}n−1, then ∫sin 3xcos xdx is.

A

cos 2x+log(sec x|+C
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B

cos 2xlog(sec x|+C
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C

cos 2xlog(sec x|+C
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D

cos 2x+log(sec x|+C
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Solution

## The correct option is C −cos 2x−log(sec x|+CWe know the reduction formula for In=∫sin nxcos xdx as : In+In−2=−2cos{(n−1)x}n−1 Now, to find In=∫sin 3xcos xdx we take n=3, thus substituting it back in the reduction formula we get I3+I1=−2cos 2x2, Here, I1=∫sinxcosxdx=ln(secx|+C, Thus we get I3=−cos 2x−ln(secx|+C So, Option c. is correct.

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