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Q.

sin3θcos3θcos2θsinθ+cosθ+cos2θ2007(sinθ)2009(cosθ)2009=

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a

12008tanθsecθ+cosecθcotθcosec2θ2008+c

b

12008(secθ+cosecθ+cotθ)2008+c

c

12009(sinθ+cosθ+tanθ)2009+c

d

12009cotθcosecθ+tanθsecθsec2θ2009+c

answer is A.

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Detailed Solution

I=sin3θcos3θcos2θsinθ+cosθ+cos2θ2007sin2θcos2θsin2007θcos2007θ=sinθcos2θcosθsin2θ1sin2θ1cosθ+1sinθ+cosθsinθ2007 put 1cosθ+1sinθ+cosθsinθ=t I=t2007dt=t20082008+c=120081cosθ+1sinθ+cosθsinθ2008+c =12008(secθ+cosecθ+cotθ)2008+c

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