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Q.

∫cos4x+1cotx−tanx dx =

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a

18 cos4x+C

b

14 cos4x+C

c

−14 cos4x+C

d

−18 cos4x+C

answer is D.

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Detailed Solution

∫2cos22x1−tan2x2tanx×2dx =∫sin2x.cos2xdx                                  =12∫sin4xdx                                  =−18cos4x+c
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