Let f(x)=x−2sinxcosx+31sin3x
⇒f′(x)=1−2cos2x+cos3x
⇒f"(x)=4sin2x−3sin3x
For maxima/minima f′(x)=0
⇒1−2(2cos2x−1)+4cos3x−3cosx=0
⇒(2cosx+3)(2cosx−3)(cosx−1)=0
cosx=2−3,23,1
x=65π,6π,0
f"(65π)=−23−3<0
f"(6π)=23−3>0
f"(0)=0
So x=65πis local maxima point
Maximum value of f(x)=f(65π)=65π+23+31
=65π+2+33
Hence this is the correct option.