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The integral I=sin2xcos2xsin5x+cos3xsin2x+sin3xcos2x+cos5x2 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

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detailed solution

Correct option is D

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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