Tích phân

N

nerversaynever

Sử dụng số phức
[TEX]\begin{array}{l}\cos ^n x\cos nx + ic{\rm{os}}^n x\sin n{\rm{x = }}c{\rm{os}}^n x\left( {\cos nx + i\sin nx} \right) = c{\rm{os}}^n x\left( {\cos x + i\sin x} \right)^n \\ = \left( {c{\rm{os}}^2 x + {\rm{i}}\sin x\cos x} \right)^n = \frac{1}{{2^n }}\left( {1 + \cos 2x + {\rm{i}}\sin 2x} \right)^n = \frac{1}{{2^n }}\sum\limits_{k = 0}^n {C_n^k \left( {\cos 2x + {\rm{i}}\sin 2x} \right)^k } \\ = \frac{1}{{2^n }}\sum\limits_{k = 0}^n {C_n^k \left( {c{\rm{os}}2kx + {\rm{i}}\sin 2kx} \right)} \\ \end{array}[/TEX]
so sánh phần thực 2 vế ta được
[TEX]\cos ^n x\cos nx = \frac{1}{{2^n }}\sum\limits_{k = 0}^n {C_n^k c{\rm{os}}2kx} [/TEX]
suy ra
[TEX]I = \frac{1}{{2^n }}\int\limits_0^{\pi /2} {\sum\limits_{k = 0}^n {C_n^k c{\rm{os}}2kx} dx = } \frac{1}{{2^n }}\left. {\left( {x + \sum\limits_{k = 1}^n {\frac{{C_n^k \sin 2kx}}{{2k}}} } \right)} \right|_0^{\pi /2} = \frac{\pi }{{2^{n + 1} }}[/TEX]
 
You must log in or register to reply here.
Top Bottom