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∫cosec2x−2005cos2005xdx

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a

cotxcosx2005+C

b

tanxcosx2005+C

c

tanxcotx2005+C

d

−cotxcosx2005+C

answer is D.

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Detailed Solution

I=∫(cos⁡x)−2005cos⁡ec2xdx−2005∫dxcos2005⁡xI=(cos⁡x)−2005(−cot⁡x)−∫(−2005)(cos⁡x)−2006(−sin⁡x)(−cot⁡x)dx−2005∫dxcos2005⁡xI=-cot⁡x(cos⁡x)2005+C

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