Evaluate:∫sin6x+cos6x+3sin2xcos2xdx
x+c
32sin2x+c
-32cos2x+c
13sin3x–cos3x+c
13sin3x+cos3x+c
Evaluating the integral
∫sin6x+cos6x+3sin2xcos2xdx
⇒∫sin2x3+cos2x3+3sin2xcos2xdx⇒∫sin2x+cos2x+sin4x+cos4x-sin2xcos2x+3sin2xcos2xdx∵a3+b3=a2+b2(a2+b2-ab)⇒∫sin4x+cos4x+2sin2xcos2xdx∵sin2x+cos2x=1⇒∫sin2x+cos2x2dx∵(a+b)2=a2+b2+2ab⇒∫dx=x+ccisanintegratingconstsnt
Hence, option A is the correct answer.