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Q.

The integral I=∫sin2⁡xcos2⁡xsin5⁡x+cos3⁡xsin2⁡x+sin3⁡xcos2⁡x+cos5⁡x2 is equal to

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a

131−tan2⁡x+C

b

131+tan3⁡x+C

c

132+tan3⁡x+C

d

−131+tan3⁡x+C

answer is D.

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Detailed Solution

sin5⁡x+cos3⁡xsin2⁡x+sin3⁡xcos2⁡x+cos5⁡x2=sin2⁡xcos2⁡xsin3⁡x+cos3⁡x2sin2⁡x+cos2⁡x2=sin2⁡xcos2⁡xsin3⁡x+cos3⁡x2=tan2⁡xsec2⁡x1+tan3⁡x2So I=∫tan2⁡xsec2⁡x1+tan3⁡x2dx=∫t21+t32dt (t=tan⁡x)=13∫duu2u=1+t3=−13u+C=−131+tan3⁡x+C
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