= Solution
Put $s=\omega t$, so the nonzero half-cycle is $0\leq s\leq\pi$. The constant coefficient is
$$
a_0=\frac{V_0}{\pi}\int_0^\pi\sin s\,ds=\frac{2V_0}{\pi}.
$$
The <sine> coefficients satisfy
$$
b_n=\frac{V_0}{\pi}\int_0^\pi\sin s\sin ns\,ds
=\begin{cases}V_0/2,&n=1,\\0,&n\ne1,\end{cases}
$$
by <sine> <orthogonality>. For the <cosine> coefficients, the product-to-sum identity gives, when $n\ne1$,
$$
\int_0^\pi\sin s\cos ns\,ds
=\frac12\int_0^\pi[\sin((1+n)s)+\sin((1-n)s)]\,ds
=\frac{1+(-1)^n}{1-n^2}.
$$
The case $n=1$ is zero directly. Hence
$$
\boxed{a_{2m}=-\frac{2V_0}{\pi(4m^2-1)},\quad a_{2m-1}=0,\quad
b_1=\frac{V_0}{2},\quad b_n=0\ (n\ne1)}.
$$
The first five nonzero terms, counting the constant as the zero-frequency term, are
$$
\boxed{V_1(t)=\frac{V_0}{\pi}+\frac{V_0}{2}\sin\omega t
-\frac{2V_0}{3\pi}\cos2\omega t
-\frac{2V_0}{15\pi}\cos4\omega t
-\frac{2V_0}{35\pi}\cos6\omega t+\cdots}.
$$
The general coefficient formula specifies every remaining term. This is the <Fourier series of a half-wave rectified sinusoid>; the waveform is continuous at the joins, so the series gives its actual value there.
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