∫1/(1+tan^n x) dx ,(0,π/2)求详解!

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∫1/(1+tan^nx)dx,(0,π/2)求详解!∫1/(1+tan^nx)dx,(0,π/2)求详解!∫1/(1+tan^nx)dx,(0,π/2)求详解!令x=π/2-t,得∫1/(1+tan

∫1/(1+tan^n x) dx ,(0,π/2)求详解!
∫1/(1+tan^n x) dx ,(0,π/2)求详解!

∫1/(1+tan^n x) dx ,(0,π/2)求详解!
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