Maximum value of  x−sinx in [0,π]  is

f(x)=x−sinx⇒f'(x)=1−cosx=0 whencosx=1 i.e. whenx=0 . Note thatf(0)=0−sin0=0 and f(π)=π−sinπ=π . Hence maximumf(x)  is  f(π)=π .