一般化

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