∫sin8x-cos8x1-2sin2xcos2xdx=?
-12sin2x+c
12sin2x+c
12sinx+c
-12sinx+c
Explanation for the correct option:
Evaluating the given integral:
∫sin8x-cos8x1-2sin2xcos2xdx
=∫sin4x2-(cos4x)2sin2x+cos2x2-2sin2xcos2xdx∵sin2x+cos2x=1=∫sin4x+cos4xsin4x-cos4xsin4x+cos4x+2sin2xcos2x-2sin2xcos2xdx∵(a2-b2)=(a-b)(a+b)&(a+b)2=a2+b2+2ab=∫sin4x+cos4xsin4x-cos4xsin4x+cos4xdx=∫sin4x-cos4xdx=∫sin2x-cos2xsin2x+cos2xdx∵(a2-b2)=(a-b)(a+b)=∫sin2x-cos2xdx=∫1-cos2x-cos2xdx∵sin2x=1-cos2x=∫1-2cos2xdx=-∫-1+2cos2xdx=-∫2cos2x-1dx=-∫cos(2x)dx∵cos(2x)=2cos2x-1=-sin(2x)2+CCisanintegratingconstant
Therefore, ∫sin8x-cos8x1-2sin2xcos2xdx=-12sin2x+c
Hence, the correct answer is option (B).