Let

$$S[x]=\int_{0}^{T} \frac{1}{2}\left(\dot{x}^{2}-\omega^{2} x^{2}\right) \mathrm{d} t, \quad x(0)=a, \quad x(T)=b .$$

For any variation $\delta x(t)$ with $\delta x(0)=\delta x(T)=0$, show that $\delta S=0$ when $x=x_{c}$ with

$$x_{c}(t)=\frac{1}{\sin \omega T}[a \sin \omega(T-t)+b \sin \omega t] .$$

By using integration by parts, show that

$$S\left[x_{c}\right]=\left[\frac{1}{2} x_{c} \dot{x}_{c}\right]_{0}^{T}=\frac{\omega}{2 \sin \omega T}\left[\left(a^{2}+b^{2}\right) \cos \omega T-2 a b\right]$$