令x=π/2-t,得∫1/(1+tan^n x) dx ,(0,π/2)=∫1/(1+cot^n x) dx ,(0,π/2),
因为∫1/(1+tan^n x) dx ,(0,π/2)+∫1/(1+cot^n x) dx ,(0,π/2)=π/2,
所以∫1/(1+tan^n x) dx ,(0,π/2)=π/4