分部积分的应用(2)

2020-02-02 本文已影响0人洛玖言

应用(2)

$\displaystyle I_n=\int_0^{\frac\pi2}\cos^nx\sin nx\text{d}x$

Sol:
$\begin{aligned} I_n=&\int_0^{\frac\pi2}\cos^nx\sin nx\text{d}x\\ =&-\dfrac1n\int_0^{\frac\pi2}\cos^nx\text{d}\cos nx\\ =&-\dfrac1n(\cos^nx\cos nx)\bigg|_0^{\frac\pi2}+\int_0^{\frac\pi2}\cos nx\cos^{n-1}x(-\sin x)\text{d}x\\ =&\dfrac1n-\int_0^{\frac\pi2}\cos^{n-1}x\cos nx\sin x\text{d}x\\ \end{aligned}$

在等式两边加上 $I_n$ ，得到
$\begin{aligned} 2I_n=&\dfrac1n+\int_0^{\frac\pi2}\cos^nx\sin nx\text{d}x-\int_0^{\frac\pi2}\cos^{n-1}x\cos nx\sin x\text{d}x\\ =&\dfrac1n+\int_0^{\frac\pi2}\cos^{n-1}x(\cos x\sin nx-\cos nx\sin x)\text{d}x\\ =&\dfrac1n+\int_0^{\frac\pi2}\cos^{n-1}x\sin(n-1)x\text{d}x\\ =&\dfrac1n+I_{n-1} \end{aligned}$

由此得到递推公式
$\displaystyle I_n=\dfrac{1}{2}\left(\dfrac{1}{n}+I_{n-1}\right)$

按照这个递推公式得到
$\begin{aligned} I_{n-1}=&\dfrac{1}{2}\left(\dfrac{1}{n-1}+I_{n-2}\right)\\ I_{n-2}=&\dfrac{1}{2}\left(\dfrac{1}{n-2}+I_{n-3}\right)\\ I_{n-3}=&\dfrac{1}{2}\left(\dfrac{1}{n-3}+I_{n-4}\right)\\ \cdots& \end{aligned}$

按此规律可以得到
$\begin{aligned} I_n=&\dfrac12\left\{\dfrac1n+\dfrac12\left[\dfrac1{n-1}+\dfrac12\left(\dfrac{1}{n-2}+\cdots\dfrac12\int_0^{\frac\pi2}\cos x\sin x\text{d}x\right)\right]\right\}\\ =&\dfrac1{2n}+\dfrac{2}{2^2(n-1)}+\cdots+\dfrac{1}{2^n}\\ =&\dfrac{1}{2^{n+1}}\left(\frac{2}{1}+\dfrac{2^2}{2}+\dfrac{2^3}{3}+\cdots+\dfrac{2^n}{n}\right)\\ =&\dfrac{1}{2^{n+1}}\sum_{k=1}^n\dfrac{2^k}{k} \end{aligned}$

类似的有
$\displaystyle J_n=\int_0^{\frac\pi2}\cos^nx\cos nx\text{d}x$

Sol:
$\begin{aligned} J_n=&\int_0^{\frac\pi2}\cos^nx\cos nx\text{d}x\\ =&\dfrac{1}{n}\int_0^{\frac\pi2}\cos^n\text{d}\sin nx\\ =&\dfrac1n(\cos^nx\sin nx)\bigg|_0^{\frac\pi2}-\dfrac1n\int_0^{\frac\pi2}\sin nx\cdot n\cos^{n-1}x(-\sin x)\text{d}x\\ =&\int_0^{\frac\pi2}\cos^{n-1}x\sin nx\sin x\text{d}x \end{aligned}$

两边同时加上 $J_n$

$\begin{aligned} 2J_n=&\int_0^{\frac\pi2}\cos^{n-1}x\sin nx\sin x\text{d}x+\int_0^{\frac\pi2}\cos^nx\cos nx\text{d}x\\ =&\int_0^{\frac\pi2}\cos^{n-1}x(\sin nx\sin x+\cos n\cos nx)\text{d}x\\ =&\int_0^{\frac\pi2}\cos^{n-1}x\cos(n-1)x\text{d}x=J_{n-1} \end{aligned}$

由此得到递推公式
$\displaystyle J_n=\dfrac{1}{2}J_{n-1}$

因此积分的结果为
$\begin{aligned} J_n=&\int_0^{\frac\pi2}\cos^nx\cos nx\text{d}x\\ =&\dfrac12\cdot\dfrac12\cdots\dfrac12\int_0^{\frac\pi2}\cos^2x\text{d}x\\ =&\dfrac{1}{2^{n-1}}\cdot\dfrac{\pi}{4}\\ =&\dfrac{\pi}{2^{n+1}} \end{aligned}$

分部积分的应用(2)

应用(2)

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