# Quod Erat Demonstrandum

## 2015/11/02

### 一個冇咩用嘅rule

Filed under: NSS,Pure Mathematics — johnmayhk @ 1:53 下午
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$u=u(x)$$v=v(x)$ 二次可導，若 $\frac{d^2u}{dx^2}=au$$\frac{d^2v}{dx^2}=bv$ 其中 $a,b$ 為常數；則

$\int uvdx=\frac{1}{b-a}(u\frac{dv}{dx}-v\frac{du}{dx})+C$

e.g.1

$\int e^x\sin xdx=?$

$u=e^x \Rightarrow \frac{d^2u}{dx^2}=e^x$　　故　$a=1$

$v=\sin x \Rightarrow \frac{d^2v}{dx^2}=-\sin x$　　故　$b=-1$

$\int e^x\sin xdx=\frac{1}{-1-1}(e^x\cos x-\sin xe^x)+C=\frac{1}{2}e^x(\sin x-\cos x)+C$

e.g.2

$\int e^{3x}\cos 4xdx=?$

$u=e^{3x} \Rightarrow \frac{d^2u}{dx^2}=9e^{3x}$　　故　$a=9$

$v=\cos 4x \Rightarrow \frac{d^2v}{dx^2}=-16\cos 4x$　　故　$b=-16$

$\int e^{3x}\cos 4xdx$

$=\frac{1}{-16-6}(e^{3x}(-4\sin 4x)-\cos 4x(3e^{3x}))+C=\frac{1}{25}e^{3x}(4\sin 4x+3\cos 4x)+C$

$\int uvdx$

$=\frac{1}{b}\int u\frac{d^2v}{dx^2}dx$

$=\frac{1}{b}\int ud(\frac{dv}{dx})$

$=\frac{1}{b}(u\frac{dv}{dx}-\int \frac{dv}{dx}du)$

$=\frac{1}{b}(u\frac{dv}{dx}-\int \frac{dv}{dx}\frac{du}{dx}dx)$

$=\frac{1}{b}(u\frac{dv}{dx}-\int \frac{du}{dx}dv)$

$=\frac{1}{b}(u\frac{dv}{dx}-v\frac{du}{dx}+\int vd(\frac{du}{dx}))$

$=\frac{1}{b}(u\frac{dv}{dx}-v\frac{du}{dx}+\int v(\frac{d^2u}{dx^2})dx)$

$=\frac{1}{b}(u\frac{dv}{dx}-v\frac{du}{dx}+a\int uvdx)$

$\int uvdx=\frac{1}{b-a}(u\frac{dv}{dx}-v\frac{du}{dx})+C$

https://johnmayhk.wordpress.com/2012/02/07/integration-by-parts/