= Solution
For an admissible <variation> $\eta$ with $\eta(0)=\eta(L)=0$,
$$
\delta U=\int_0^L(EI\theta'\eta'-P\sin\theta\,\eta)\,ds=-\int_0^L(EI\theta''+P\sin\theta)\eta\,ds.
$$
Stationarity for every <variation> therefore gives the <Euler-Lagrange equation>
$$
\boxed{EI\theta''+P\sin\theta=0.}
$$
The first energy term is <bending energy>: the <curvature> of an inextensible planar beam is $\theta_s$. The shortening relative to the straight reference is $\int(1-\cos\theta)\,ds$. A prescribed compressive <dead load> does positive work on that shortening, so its potential-energy contribution is minus $P$ times that integral. A <minimizer> must satisfy the stationary equation, but a stationary solution need not be a <minimizer>.
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