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Q.

∫dxcosx cos2x=

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a

cos−1(tanx)+c

b

cos(tan−1x)+c

c

sin−1(tanx)+c

d

tan(sin−1x)+c

answer is C.

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Detailed Solution

∫1cosxcos2x−sin2xdx                  =∫sec2x1−tan2xdx                        tanx=t                   =∫11−t2dt                   =sin−1t+c                   =sin−1(tanx)+c
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