Ngô Quốc Anh

October 4, 2007

1 bài tích phân xác định

Filed under: Các Bài Tập Nhỏ, Giải Tích 2 — Ngô Quốc Anh @ 16:50

Tính tích phân I_n\equiv\int_0^{\frac {\pi}{2}}\cos^nx\sin (n + 2)x\ \mathrm {dx} = A + B.

Lời giải.

I_n\equiv\int_0^{\frac {\pi}{2}}\cos^nx\sin (n + 2)x\ \mathrm {dx} = A + B, where

A\equiv\int_0^{\frac {\pi}{2}}\cos^nx\sin x\cos (n + 1)x\ \mathrm {dx} and B\equiv\int_0^{\frac {\pi}{2}}\cos^{n + 1}x\sin (n + 1)x\ \mathrm {dx}\ .

\boxed {\ \begin{array}{ccc} u(x) = \cos (n + 1)x & \Longrightarrow & u'(x) = - (n + 1)\sin (n + 1)x\ . \\  <br>  \\  <br> v'(x) = \cos^nx\sin x & \Longrightarrow & v(x) = - \frac {1}{n + 1}\cdot\cos^{n + 1}x\ .\end{array}\ }\right\| 

A = \left[ - \frac {1}{n + 1}\cdot \cos^{n + 1}x\cos (n + 1)x\right]_0^{\frac {\pi}{2}} - B 

A + B = \left[ - \frac {1}{n + 1}\cdot \cos^{n + 1}x\cos (n + 1)x\right]_0^{\frac {\pi}{2}} 

 \boxed {\ I_n = \frac {1}{n + 1}\ }\ .

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