Merge remote-tracking branch 'origin/develop' into develop

素材/整合素材/高数素材/常用不定积分表.md

@@ -1,47 +1,47 @@
 一、基本初等函数积分
- $\int 0 \, dx =C$
- $\int k \, dx = kx + C$ （ k  为常数）
- $\int x^\mu \, dx = \frac{x^{\mu+1}}{\mu+1} + C$ （ $\mu \neq -1$ ）
- $\int \frac{1}{x} \, dx = \ln|x| + C$ 
-$\int a^x \, dx = \frac{a^x}{\ln a} + C$ （ $a>0,a\neq1$ ）
- $\int e^x \, dx = e^x + C$ 
+ $\displaystyle\int 0 \, dx =C$
+ $\displaystyle\int k \, dx = kx + C$ （ k  为常数）
+ $\displaystyle\int x^\mu \, dx = \frac{x^{\mu+1}}{\mu+1} + C$ （ $\mu \neq -1$ ）
+ $\displaystyle\int \frac{1}{x} \, dx = \ln|x| + C$ 
+$\displaystyle\int a^x \, dx = \frac{a^x}{\ln a} + C$ （ $a>0,a\neq1$ ）
+ $\displaystyle\int e^x \, dx = e^x + C$ 
 二、三角函数积分
- $\int \sin x \, dx = -\cos x + C$ 
- $\int \cos x \, dx = \sin x + C$ 
- $\int \tan x \, dx = -\ln|\cos x| + C$
- $\int \cot x \, dx = \ln|\sin x| + C$
- $\int \sec x \, dx = \ln|\sec x + \tan x| + C$ 
- $\int \csc x \, dx = \ln|\csc x - \cot x| + C$ 
- $\int \sec^2 x \, dx = \tan x + C$ 
- $\int \csc^2 x \, dx = -\cot x + C$ 
- $\int \sec x \tan x \, dx = \sec x + C$ 
- $\int \csc x \cot x \, dx = -\csc x + C$ 
- $\int \sin^2 x \, dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$ 
- $\int \cos^2 x \, dx = \frac{x}{2} + \frac{\sin 2x}{4} + C$ 
+ $\displaystyle\int \sin x \, dx = -\cos x + C$ 
+ $\displaystyle\int \cos x \, dx = \sin x + C$ 
+ $\displaystyle\int \tan x \, dx = -\ln|\cos x| + C$
+ $\displaystyle\int \cot x \, dx = \ln|\sin x| + C$
+ $\displaystyle\int \sec x \, dx = \ln|\sec x + \tan x| + C$ 
+ $\displaystyle\int \csc x \, dx = \ln|\csc x - \cot x| + C$ 
+ $\displaystyle\int \sec^2 x \, dx = \tan x + C$ 
+ $\displaystyle\int \csc^2 x \, dx = -\cot x + C$ 
+ $\displaystyle\int \sec x \tan x \, dx = \sec x + C$ 
+ $\displaystyle\int \csc x \cot x \, dx = -\csc x + C$ 
+ $\displaystyle\int \sin^2 x \, dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$ 
+ $\displaystyle\int \cos^2 x \, dx = \frac{x}{2} + \frac{\sin 2x}{4} + C$ 
 三、反三角函数积分
- $\int \arcsin x \, dx = x\arcsin x + \sqrt{1-x^2} + C$ 
- $\int \arccos x \, dx = x\arccos x - \sqrt{1-x^2} + C$ 
- $\int \arctan x \, dx = x\arctan x - \frac{1}{2}\ln(1+x^2) + C$ 
- $\int \text{arccot } x \, dx = x\text{arccot } x + \frac{1}{2}\ln(1+x^2) + C$ 
+ $\displaystyle\int \arcsin x \, dx = x\arcsin x + \sqrt{1-x^2} + C$ 
+ $\displaystyle\int \arccos x \, dx = x\arccos x - \sqrt{1-x^2} + C$ 
+ $\displaystyle\int \arctan x \, dx = x\arctan x - \frac{1}{2}\ln(1+x^2) + C$ 
+ $\displaystyle\int \text{arccot } x \, dx = x\text{arccot } x + \frac{1}{2}\ln(1+x^2) + C$ 
 四、含根式的积分（ a>0 ）
- $\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \arcsin\frac{x}{a} + C$ 
- $\int \frac{1}{\sqrt{x^2 + a^2}} \, dx = \ln\left(x + \sqrt{x^2 + a^2}\right) + C$ 
- $\int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \ln\left|x + \sqrt{x^2 - a^2}\right| + C$ 
- $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C$ 
- $\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left(x + \sqrt{x^2 + a^2}\right) + C$ 
- $\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C$ 
+ $\displaystyle\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \arcsin\frac{x}{a} + C$ 
+ $\displaystyle\int \frac{1}{\sqrt{x^2 + a^2}} \, dx = \ln\left(x + \sqrt{x^2 + a^2}\right) + C$ 
+ $\displaystyle\int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \ln\left|x + \sqrt{x^2 - a^2}\right| + C$ 
+ $\displaystyle\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C$ 
+ $\displaystyle\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left(x + \sqrt{x^2 + a^2}\right) + C$ 
+ $\displaystyle\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C$ 
 五、含分式的积分（ $a\neq0$ ）
- $\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a}\arctan\frac{x}{a} + C$ 
- $\int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a}\ln\left|\frac{x - a}{x + a}\right| + C$ 
- $\int \frac{1}{ax + b} \, dx = \frac{1}{a}\ln|ax + b| + C$ 
+ $\displaystyle\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a}\arctan\frac{x}{a} + C$ 
+ $\displaystyle\int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a}\ln\left|\frac{x - a}{x + a}\right| + C$ 
+ $\displaystyle\int \frac{1}{ax + b} \, dx = \frac{1}{a}\ln|ax + b| + C$ 
 六、指数与对数结合积分
- $\int x e^x \, dx = (x-1)e^x + C$ 
-$\int x \ln x \, dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C$ 
- $\int e^x \sin x \, dx = \frac{e^x}{2}(\sin x - \cos x) + C$ 
- $\int e^x \cos x \, dx = \frac{e^x}{2}(\sin x + \cos x) + C$ 
+ $\displaystyle\int x e^x \, dx = (x-1)e^x + C$ 
+$\displaystyle\int x \ln x \, dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C$ 
+ $\displaystyle\int e^x \sin x \, dx = \frac{e^x}{2}(\sin x - \cos x) + C$ 
+ $\displaystyle\int e^x \cos x \, dx = \frac{e^x}{2}(\sin x + \cos x) + C$ 
 七、常用凑微分积分
- $\int \frac{1}{\sqrt{1 - x^2}} \, dx = \arcsin x + C = -\arccos x + C$ 
- $\int \frac{1}{1 + x^2} \, dx = \arctan x + C = -\text{arccot } x + C$ 
- $\int \frac{1}{x\ln x} \, dx = \ln|\ln x| + C$ 
- $\int \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx = 2e^{\sqrt{x}} + C$ 
- $\int \frac{\sin\sqrt{x}}{\sqrt{x}} \, dx = -2\cos\sqrt{x} + C$ 
+ $\displaystyle\int \frac{1}{\sqrt{1 - x^2}} \, dx = \arcsin x + C = -\arccos x + C$ 
+ $\displaystyle\int \frac{1}{1 + x^2} \, dx = \arctan x + C = -\text{arccot } x + C$ 
+ $\displaystyle\int \frac{1}{x\ln x} \, dx = \ln|\ln x| + C$ 
+ $\displaystyle\int \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx = 2e^{\sqrt{x}} + C$ 
+ $\displaystyle\int \frac{\sin\sqrt{x}}{\sqrt{x}} \, dx = -2\cos\sqrt{x} + C$ 

素材/整合素材/高数素材/积分题目.md

@@ -201,4 +201,35 @@ I_n &= \int_0^{\frac{\pi}{4}} \sec^n x  \, \mathrm{d}x = \int_0^{\frac{\pi}{4}}
 >用分部积分法得 $$I=xf(x)|^1_0-\int_0^1xf'(x)\text dx=-\int_0^1x\arctan(x^2-1)\text dx.$$
 >令 $u=x^2-1$，则 $\text du=2x\text dx$，于是 $$I=-\frac{1}{2}\int_{-1}^0\arctan u\text du=-\frac{1}{2}((u\arctan u)|_{-1}^0-\frac{1}{2}\ln(1+u^2)|_{-1}^0)=\frac{\pi}{8}-\frac{1}{4}\ln2.$$
 
+>[!example] 例题
+>设 $\displaystyle a_n=\int_0^1x(1-x)^n\text dx,n=1,2,\cdots$.
+>（1）求级数 $\displaystyle\sum_{n=1}^\infty a_n$；
+>（2）设常数 $\lambda\gt0$，试讨论级数 $\displaystyle\sum_{n=1}^\infty \lambda^na_n$ 的敛散性.
 
+>[!note] 解析
+>（1）**解1：**$$\begin{aligned}
+>a_n&=-\int_0^1x(1-x)^n\text d(1-x)\\
+>&=-\frac{1}{n+1}\int_0^1x\text d(1-x)^{n+1}\\
+>&=-\frac{1}{n+1}\left([x(1-x)^{n+1}]_0^1-\int_0^1(1-x)^{n+1}\text dx\right)\\
+>&=-\frac{1}{n+1}\int_0^1(1-x)^{n+1}\text d(1-x)\\
+>&=-\frac{1}{n+1}\left[\frac{(1-x)^{n+2}}{n+2}\right]_0^1\\
+>&=\frac{1}{(n+1)(n+2)}\\
+>&=\frac{1}{n+1}-\frac{1}{n+2}.
+>\end{aligned}$$
+>于是$$\sum_{k=1}^n a_k = \sum_{k=1}^n \left( \frac{1}{k+1} - \frac{1}{k+2} \right) = \frac{1}{2} - \frac{1}{n+2}.$$
+>因此$$\sum_{n=1}^\infty a_n = \lim_{n \to \infty} \left( \frac{1}{2} - \frac{1}{n+2} \right) = \frac{1}{2}.$$
+>**解2：** 令 $t=1-x$，则
+>$$\begin{aligned}
+a_n &= \int_0^1 x (1-x)^n \, dx = \int_0^1 (1-t) t^n \, dt \\
+&= \int_0^1 (t^n - t^{n+1}) \, dt = \left[ \frac{t^{n+1}}{n+1} - \frac{t^{n+2}}{n+2} \right]_0^1 \\
+&= \frac{1}{n+1} - \frac{1}{n+2}.
+>\end{aligned}$$
+>后同解1
+>（2）设 $\displaystyle b_n = \lambda^n a_n = \lambda^n \left( \frac{1}{n+1} - \frac{1}{n+2} \right)$。考虑比值判别法：$$\lim_{n \to \infty} \frac{b_{n+1}}{b_n} = \lim_{n \to \infty} \frac{\lambda^{n+1} \left( \frac{1}{n+2} - \frac{1}{n+3} \right)}{\lambda^n \left( \frac{1}{n+1} - \frac{1}{n+2} \right)} = \lambda \lim_{n \to \infty} \frac{ \frac{1}{(n+2)(n+3)} }{ \frac{1}{(n+1)(n+2)} } = \lambda \lim_{n \to \infty} \frac{n+1}{n+3} = \lambda.$$
+>故当 $0 < \lambda < 1$ 时，级数收敛；当 $\lambda > 1$ 时，级数发散。
+当 $\lambda = 1$ 时，$\displaystyle b_n = \frac{1}{n+1} - \frac{1}{n+2}$，由 (1) 知级数收敛。
+综上，当 $0 < \lambda \le 1$ 时，级数 $\displaystyle \sum_{n=1}^\infty \lambda^n a_n$ 收敛；当 $\lambda > 1$ 时，级数发散.
+
+>[!summary] 题后总结
+>第一问解1是利用第一类换元法和分部积分法，比较常规，也相对容易想到；解2利用第二类换元法，大大简化了计算，可以作为经验积累下来。为什么用这种换元呢？观察一下特征：一个相对复杂的式子 $(1-x)$ 上有一个 $n$ 次方，变得更加复杂，而简单的式子 $x$ 是一次的，更加简单。为了“调和”两个式子的复杂度，我们把复杂式子代换掉，这样就能更加方便地进行积分了。
+