= 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 are
$$
\boxed{V_2(t)=\frac{V_0}{\pi}-\frac{V_0}{2}\sin\omega t
-\frac{2V_0}{3\pi}\cos2\omega t
-\frac{2V_0}{15\pi}\cos4\omega t
-\frac{2V_0}{35\pi}\cos6\omega t+\cdots}.
$$
This 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.
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