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Q.

I=∫sin10⁡x−cos8⁡xsin2⁡x+sin8⁡xcos2⁡x−cos10⁡x1−2sin2⁡xcos2⁡xdx is equal to

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a

12sin⁡2x+C

b

−12sin⁡2x+C

c

−12sin⁡x+C

d

−sin2⁡x+C

answer is B.

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Detailed Solution

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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