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I=sin10xcos8xsin2x+sin8xcos2xcos10x12sin2xcos2xdx is equal to

 

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a
12sin⁡2x+C
b
−12sin⁡2x+C
c
−12sin⁡x+C
d
−sin2⁡x+C

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detailed solution

Correct option is B

sin10⁡x−cos8⁡xsin2⁡x+sin8⁡xcos2⁡x−cos10⁡x=sin2⁡xsin8⁡x−cos8⁡x+cos2⁡xsin8⁡x−cos8⁡x=sin2⁡x+cos2⁡xsin8⁡x−cos8⁡x=sin2⁡x−cos2⁡xsin2⁡x+cos2⁡xsin4⁡x+cos4⁡x=sin2⁡x−cos2⁡x1−2sin2⁡xcos2⁡x=−cos⁡2x1−2sin2⁡xcos2⁡xI=−∫cos⁡2xdx=−sin⁡2x2+C


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