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∫1cos2xsin4xdx=

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a

cotx−2tanx−cot3x3+C

b

cotx+2tanx−cot3x3+c

c

cotx+2tanx+tan3x3+c

d

tanx−2cotx−cot3x3+c

answer is D.

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Detailed Solution

I=∫1cos2xsin4xdx=∫sin2x+cos2x2cos2xsin4xdx=∫sin4⁡x+2sin2⁡xcos2⁡x+cos4⁡xcos2⁡xsin4⁡xdx=∫sec2⁡x+2cosec2⁡x+cot2⁡xcosec2⁡x=tan⁡x−2cot⁡x−cot3⁡x3+c

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