积分

三角函数定积分

-1次：

$$\begin{array}{rl}& {\int }_0^{\pi /2}\frac{1}{\sin x+\cos x}dx\\ & =\frac{1}{\sqrt{2}}{\int }_0^{\pi /2}\sec \left(x-\frac{\pi }{4}\right)dx\\ & =\frac{1}{\sqrt{2}}\ln \left|\sec \left(x-\frac{\pi }{4}\right)+\tan \left(x-\frac{\pi }{4}\right)\right|{|}_0^{\pi /2}\\ & =\frac{1}{\sqrt{2}}\ln \frac{\sqrt{2}+1}{\sqrt{2}-1}\end{array}$$

一次：

$${\int }_0^{\pi /2}\sin xdx={\int }_0^{\pi /2}\cos xdx=1$$ $${\int }_0^{\pi /4}\sin xdx=\frac{\sqrt{2}}{2};{\int }_0^{\pi /4}\cos xdx=1-\frac{\sqrt{2}}{2};{\int }_0^{\pi /4}\tan xdx=\frac{1}{2}\ln 2;$$

二次：

$${\int }_0^{\pi /2}{\sin }^{2}xdx={\int }_0^{\pi /2}{\cos }^{2}xdx=\frac{\pi }{4}$$ $${\int }_0^{\pi /2}\sin x\cos xdx=\frac{1}{2}{\sin }^{2}x{|}_0^{\pi /2}=\frac{1}{2}$$

三次：

$${\int }_0^{\pi /2}{\sin }^{2}x\cos xdx=\frac{1}{3}{\sin }^{3}x{|}_0^{\pi /2}=\frac{1}{3}$$ $${\int }_0^{\pi /2}{\cos }^{2}x\sin xdx=-\frac{1}{3}{\cos }^{3}x{|}_0^{\pi /2}=\frac{1}{3}$$ $${\int }_0^{\pi /2}{\sin }^{3}xdx={\int }_0^{\pi /2}{\cos }^{3}xdx=\frac{2}{3}\times 1=\frac{2}{3}$$ <font color="#000000">\int _{0}^{\pi/2}\sin ^{1/3}x\cos ^{5/3}x+\sin ^{-1/3}x\cos ^{7/3}x \, dx </font> $${\int }_0^{+\infty }\frac{u^2}{1+u^4}du$$ $${\int }_0^{\sqrt{\pi /4}}{x}^2(1-\tan x^2){\sec }^2{x}^{2}2xdx$$ $$\begin{array}{rl}& \int {\sec }^{3}xdx\\ & =\int {\sec }^{2}x\sec xdx\\ & =\int \sec xd\tan x\\ & =\sec x\tan x-\int \tan xd\sec x\\ & =\sec x\tan x-\int {\tan }^{2}x\sec xdx\\ & =\sec x\tan x-\int ({\sec }^{2}x-1)\sec xdx\\ & =\sec x\tan x-\int {\sec }^{3}xdx+\int \sec xdx\\ & =\sec x\tan x-\int {\sec }^{3}xdx+\ln |\sec x+\tan x|\\ & =\frac{1}{2}(\sec x\tan x+\ln |\sec x+\tan x|)+C\end{array}$$ $$\begin{array}{rl}& \int {\sec }^{4}xdx\\ & =\int {\sec }^{2}x{\sec }^{2}xdx\\ & =\int {\sec }^{2}xd\tan x\\ & ={\sec }^{2}x\tan x-\int \tan xd{\sec }^{2}x\\ & ={\sec }^{2}x\tan x-2\int \tan x\sec x\sec x\tan xdx\\ & ={\sec }^{2}x\tan x-2\int ({\sec }^{2}x-1){\sec }^{2}xdx\\ & ={\sec }^{2}x\tan x-2\int {\sec }^4-{\sec }^{2}xdx\\ & ={\sec }^{2}x\tan x-2\int {\sec }^{4}dx+2\int {\sec }^{2}xdx\\ & ={\sec }^{2}x\tan x-2\int {\sec }^{4}dx+2\tan x\\ & =\frac{1}{3}({\sec }^{2}x\tan x+2\tan x)+C\end{array}$$ $$\int \tan xdx=-\ln |\cos x|$$ $$\int {\tan }^{2}xdx=\int {\sec }^{2}x-1dx=\tan x-x+C$$ $$\begin{array}{rl}& \int {\tan }^{3}dx\\ & =\int \tan x({\sec }^{2}x-1)dx\\ & =\int \tan xd\tan x-\int \tan xdx\\ & =\frac{1}{2}{\tan }^{2}x+\ln |\cos x|+C\end{array}$$ $$\begin{array}{rl}& \int {\tan }^{4}dx\\ & =\int {\tan }^{2}x{\tan }^{2}xdx\\ & =\int {\tan }^{2}x({\sec }^{2}x-1)dx\\ & =\int {\tan }^{2}xd\tan x-\int {\tan }^{2}xdx\\ & =\frac{1}{3}{\tan }^{3}x-\int ({\sec }^{2}x-1)dx\\ & =\frac{1}{3}{\tan }^{3}x-\tan x+x+C\end{array}$$ $${\int }_0^{\pi }xf(\sin x)dx=\frac{\pi }{2}{\int }_0^{\pi }f(\sin x)dx=\pi {\int }_0^{\frac{\pi }{2}}f(\sin x)dx$$

表格法简化反复分部积分

$\ln x$	$\frac{1}{x}$
$x$	$\frac{1}{2}{x}^2$

$$\int x\ln xdx=\frac{1}{2}{x}^2\ln x-\int \frac{1}{2}xdx=\frac{1}{2}{x}^2\ln x-\frac{1}{4}{x}^2$$

拉格朗日乘数法

欧拉齐次函数定理简化计算

目标函数 $F(x,y,z)$ ,约束条件 $G(x,y,z)=C$ ,其中 $F$ 与 $G$ 为齐次多项式

设 $L(x,y,z,\lambda )=F(x,y,z)+\lambda G(x,y,z)$

$$\{\begin{array}{l}{L}_x^{\prime }=F_x^{\prime }+\lambda G_x^{\prime }=0\\ L_y^{\prime }=F_y^{\prime }+\lambda G_y^{\prime }=0\\ L_z^{\prime }=F_z^{\prime }+\lambda G_z^{\prime }=0\\ L_{\lambda }^{\prime }=G-C=0\end{array}$$

可得

$$x{L}_x^{\prime }+y{L}_y^{\prime }+z{L}_z^{\prime }=0$$

代入偏导可得

$$x{F}_x^{\prime }+y{F}_y^{\prime }+z{F}_z^{\prime }+\lambda (x{G}_x^{\prime }+y{G}_y^{\prime }+z{G}_z^{\prime })=0$$

由欧拉齐次函数定理

$$x{F}_x^{\prime }+y{F}_y^{\prime }+z{F}_z^{\prime }=F,x{G}_x^{\prime }+y{G}_y^{\prime }+z{G}_z^{\prime }=G$$

故

$$F+\lambda G=F+\lambda C=0$$

故在极值处有

$$F=-\lambda C$$

$\int e^x\ln xdx$