Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 18 / d / Solution

The given identity immediately yields $V_2=V_1-V_0\sin \omega t$. Thus only the fundamental sine coefficient changes: $V_0/2$ becomes $-V_0/2$. No new integral is needed. The first five nonzero terms areThis also follows from translation of a Fourier series: a half-period delay multiplies the $n$th harmonic by $(-1{)}^n$, leaving the constant and even harmonics unchanged and reversing the only odd harmonic present.