Bài 111815

Question

Tính tích phân sau theo $n \in N : R_n = \int\limits_{0}^{\pi} \cos ^n x \cos nxdx.$

score 0 · Answer 1

Đặt $\begin{cases}u = \cos ^n x \\ dv = \cos nxdx \end{cases} \Rightarrow \begin{cases}du = - n \cos ^{n-1}x \sin xdx \\ v = \frac{1}{n} \sin nx \end{cases} $
$R_n = \frac{1}{n}(\cos ^n x \sin nx )\left| \begin{array}{l}
\pi \\
0
\end{array} \right. + \int\limits_{0}^{\pi} \cos ^{n-1} x \sin x \sin nxdx$
$= \int\limits_{0}^{\frac{\pi}{2} } \cos ^{n-1} x \sin x \sin nxdx =\frac{1}{2} \int\limits_{0}^{\pi} cos ^{n-1} x [ \cos (n-1) x - \cos (n+1) x]dx$
$=\frac{1}{2} \int\limits_{0}^{\pi} \cos ^{-1} x . \cos (n-1) xdx = \frac{1}{2} \int\limits_{0}^{\pi}\cos ^{n-1} x. \cos (n+1) xdx$
$=\frac{1}{2} R_{n-1} -\frac{1}{2} \int\limits_{0}^{\pi} \cos ^{n-1} x( \cos nx \cos x - \sin nx \sin x)dx$
$=\frac{1}{2} R_{n-1} -\frac{1}{2} \int\limits_{0}^{\pi} \cos ^n x \cos nxdx _ \frac{1}{2} \int\limits_{0}^{\pi} \cos ^{n-1} x \sin x \sin nxdx$
$=\frac{1}{2} R_{n-1} -\frac{1}{2} R_n +\frac{1}{2} R_n = \frac{1}{2}R_{n-1}$
$\Rightarrow R_n =\frac{1}{2} R_{n-1} =\frac{1}{2^2}R_{n-2} -\frac{1}{2^3}R_{n-3} = ...= \frac{1}{2^{n-1}}R_1.$
Chủ ý rằng : $ R_1 = \int\limits_{0}^{\pi} \cos ^2 xdx = \frac{1}{2} \int\limits_{0}^{\pi}(1 +\cos 2x)dx = \left ( \frac{x}{2}+\frac{\sin 2x}{4} \right )\left| \begin{array}{l}
\pi \\
0
\end{array} \right. =\frac{\pi}{2}$
Vậy $ R_n =\frac{\pi}{2^n}.$

Bài 111815

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Bài 111815

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