= Solution
Put $s_n(x)=\sin(q_nx)$ and $s_m(x)=\sin(q_mx)$. Their <differential equations> are $-s_n''=q_n^2s_n$ and $-s_m''=q_m^2s_m$. Subtract the two cross-multiplied equations and use <integration by parts>:
$$
(q_n^2-q_m^2)\int_0^1s_ns_m\,dx=[s_ns_m'-s_ms_n']_0^1.
$$
The lower boundary term is zero because both functions vanish there. At the upper endpoint, the <Robin boundary condition> gives $s_j'(1)=s_j(1)/\alpha$, so
$$
s_n(1)s_m'(1)-s_m(1)s_n'(1)=\frac{s_n(1)s_m(1)}\alpha-\frac{s_m(1)s_n(1)}\alpha=0.
$$
Since distinct positive roots have $q_n^2\ne q_m^2$, we conclude
$$
\boxed{\int_0^1\sin(q_nx)\sin(q_mx)\,dx=0,\qquad n\ne m.}
$$
This <orthogonality> holds for every $\alpha>0$, even when the sine family requires the additional nonpositive mode for completeness as explained in part (c).
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