Now, If
f(2a−x)=f(x), then
∫2a0f(x)dx=2∫a0f(x)dx
⟹I=2∫π/20sin6xdx=I1(say)
I=2∫π/20sin6(π2−x)dx
=2∫π/20cos6xdx=I2
Hence, 2I=2∫π/20(sin6x+cos6x)dx
⟹I=∫π/20[(sin2x+cos2x)3−3sin2xcos2x(sin2x+cos2x)]dx
⟹I=∫π/20(1−3sin2xcos2x)dx
⟹I=∫π/20(1−34(4sin2xcos2x))dx
⟹I=∫π/20(1−34sin22x)dx
⟹I=∫π/20(1−34.1−cos4x2)dx
⟹I=∫π/20(58+38cos4x)dx
⟹I=58[x]π/20+332[sin4x]π/20
⟹I=5π16
Hence, answer is option-(A).