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$$\begin{array}{rl}\sum _{0<n}\frac{\cos nx}{n}& =\sum _{0<n}\frac{1}{n}\mathrm{Re}(e^{inx})\\ & =\mathrm{Re}(-\ln (1-e^{ix}))\\ & =-\mathrm{Re}(\ln (\frac{e^{\frac{ix}{2}}-e^{-\frac{ix}{2}}}{2i}\cdot \frac{2e^{\frac{ix}{2}}}{i}))\\ & =-\mathrm{Re}(\ln (2\sin \frac{x}{2}\cdot \frac{e^{\frac{ix}{2}}}{i}))\\ & =-\ln (|2\sin \frac{x}{2}|\cdot \left|\frac{e^{\frac{ix}{2}}}{i}\right|)\\ & =-\ln |2\sin \frac{x}{2}|\end{array}$$

let$t$as$\cos x$in theorem1, and use equality$T_n(\cos x)=\cos \left(nx\right)$and${\displaystyle {\sin }^2\frac{x}{2}=\frac{1-\cos x}{2}}$

$$\begin{array}{rl}{\int }_0^{\pi /2}{\tan }^x\theta d\theta & ={\int }_0^{\pi /2}{\sin }^x\theta {\cos }^{-x}\theta d\theta \\ & ={\int }_0^{\pi /2}{\sin }^{2\cdot \frac{1+x}{2}-1}\theta {\cos }^{2\cdot \frac{1-x}{2}-1}\theta d\theta \\ & =\frac{1}{2}\mathrm{B}(\frac{1+x}{2},\frac{1-x}{2})\\ & =\frac{\mathrm{\Gamma }(\frac{1+x}{2})\mathrm{\Gamma }(\frac{1-x}{2})}{2\mathrm{\Gamma }(1)}\\ & =\frac{\pi }{2\sin \frac{\pi (1+x)}{2}}\\ & =\frac{\pi }{2}\csc (\frac{\pi }{2}+\frac{\pi x}{2})\\ & =\frac{\pi }{2}\sec \frac{\pi x}{2}\end{array}$$