微积分——巧解不定积分和定积分问题

1. 知识归纳：

1.1 定积分性质

以下结论在$f\left(x\right),g\left(x\right),\left|f\left(x\right)\right|,$ 在 $\left[a,b\right]\left[a,c\right]\left[c,b\right]$ 可积，且 $a<c<b$

1. (线性性质) ${\int }_a^b\left[\alpha f\left(x\right)\pm \beta g\left(x\right)\right]dx=\alpha {\int }_a^{b}f\left(x\right)dx\pm \beta {\int }_a^{b}g\left(x\right)dx\left(\alpha \beta \in R\right)$
2. (非负性质) $f\left(x\right)$ 在 $\left[a,b\right]$ 上非负可积，则${\int }_a^{b}f\left(x\right)dx\ge 0$
3. (单调性)若 $f\left(x\right)\ge g\left(x\right)\left(x\in \left[a,b\right]\right)$，则 ${\int }_a^{b}f\left(x\right)dx\ge {\int }_a^{b}g\left(x\right)dx$
4. 若 $m\le f\left(x\right)\le M\left(x\in \left[a,b\right]\right),m\left(b-a\right)\le {\int }_a^{b}f\left(x\right)dx\le M\left(b-a\right)$
5. $\left|{\int }_a^{b}f\left(x\right)dx\right|\le {\int }_a^b\left|f\left(x\right)\right|dx$
6. (区间可加性) ${\int }_a^{b}f\left(x\right)dx={\int }_a^{c}f\left(x\right)dx+{\int }_c^{b}f\left(x\right)dx$
7. (积分中值定理) ${\int }_a^{b}f\left(x\right)g\left(x\right)dx=f\left(\xi \right){\int }_a^{b}g\left(x\right)dx$ 推论：令 $g\left(x\right)=1{\int }_a^{b}f\left(x\right)dx=f\left(\xi \right)\left(b-a\right)$,$\left(积分中值：\frac{{\int }_a^{b}f\left(x\right)dx}{\left(b-a\right)}\right)$
8. $\begin{array}{l}Cauchy-Schwartz不等式:{\left({\int }_a^{b}f\left(x\right)g\left(x\right)dx\right)}^2\le {\int }_a^b{f}^2\left(x\right)dx{\int }_a^b{g}^2\left(x\right)dx\left(竞赛必记\right)\\ Minkowwski不等式:{\left({\int }_a^b{\left[f\left(x\right)+g\left(x\right)\right]}^{2}dx\right)}^{\frac{1}{2}}\le {\left({\int }_a^b{f}^2\left(x\right)dx\right)}^{\frac{1}{2}}+{\left({\int }_a^b{g}^2\left(x\right)dx\right)}^{\frac{1}{2}}\end{array}$
9. 变上下限积分求导：一般，若 $u\left(x\right),v\left(x\right)$ ，是可微函数，$f$ 是连续函数，则有：$\frac{d}{dx}\left({\int }_{v\left(x\right)}^{u\left(x\right)}f\left(t\right)dt\right)=f(u(x))u^{\prime }(x)-f(v(x))v^{\prime }(x)$

1.2 计算技巧公式：

1. $f(x)$ 奇函数，则 ${\int }_{-a}^{a}f\left(x\right)dx=0$

2. $f(x)$ 偶函数，则 ${\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$

3. $f(x)$ 周期函数 $l$ 函数，则 ${\int }_a^{a+l}f\left(x\right)dx={\int }_0^{l}f\left(x\right)dx={\int }_{-\frac{l}{2}}^{\frac{l}{2}}f\left(x\right)dx$

4. (1). $f(x)$ 在 $[0,a]$ ，$\left[0,a\right],{\int }_0^{a}f\left(x\right)dx={\int }_0^{\frac{a}{2}}f\left(x\right)dx+{\int }_0^{\frac{a}{2}}f\left(a-x\right)dx$

   ​ ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx$ (这个公式用的比较多)

   (2). $f(x)$ 在 $[-a,a]$ ，${\int }_{-a}^{a}f\left(x\right)dx={\int }_0^a\left[f\left(x\right)+f\left(-x\right)\right]dx$

   ​ 补充：${\int }_0^{\pi }xf\left(\sin x\right)dx=\frac{\pi }{2}{\int }_0^{\pi }f\left(\sin x\right)dx$

2. 不定积分

例题1：$\int \frac{1}{1+x^4}dx$

$\begin{array}{l}=\frac{1}{2}\int \frac{1+x^2}{1+x^4}-\frac{x^2-1}{1+x^4}dx=\frac{1}{2}\int \frac{d\left(x-\frac{1}{x}\right)}{{\left(x-\frac{1}{x}\right)}^2+2}-\frac{1}{2}\int \frac{d\left(x+\frac{1}{x}\right)}{{\left(x+\frac{1}{x}\right)}^2-2}\\ \\ =\frac{1}{2\sqrt{2}}\arctan \frac{1}{\sqrt{2}}\left(x-\frac{1}{x}\right)-\frac{1}{4\sqrt{2}}\ln \left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+c\left(x\ne 0\right)\end{array}$

例题2：$\int \frac{dx}{{\sin }^{3}x\cos x}$

$\begin{array}{l}=\int \frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{3}x\cos x}dx=\int \frac{d\tan x}{\tan x}+\int \frac{d\sin x}{{\sin }^{3}x}\\ \\ =\ln \left|\tan x\right|-\frac{1}{2{\sin }^{2}x}+C\end{array}$

例题3：$\int \frac{x+\sin x}{1+\cos x}dx$

$\begin{array}{l}=\int \frac{x}{1+\cos x}dx+\int \frac{\sin x}{1+\cos x}dx=\int \frac{x}{{\cos }^2\frac{x}{2}}d\frac{x}{2}+\int \frac{\sin x}{1+\cos x}dx\\ \\ =x\tan \frac{x}{2}-\int \tan \frac{x}{2}dx+\int \tan \frac{x}{2}dx=x\tan \frac{x}{2}+C\end{array}$

例题4： $\int \sqrt{\frac{1-x}{1+x}}\frac{dx}{x}$ 令 $t=\sqrt{\frac{1-x}{1+x}},x=\frac{1-t^2}{1+t^2},dx=\frac{-4t}{{\left(1+t^2\right)}^2}dt$

$=\int t\cdot \frac{\frac{-4t}{{\left(1+t^2\right)}^2}}{\frac{1-t^2}{1+t^2}}dt=\int \frac{-4t^2}{\left(1+t^2\right)\left(1-t^2\right)}dt=2\int \frac{1}{1+t^2}-\frac{1}{1-t^2}dt$

$=2\arctan t+\ln \left|\frac{t-1}{t+1}\right|=2\arctan \sqrt{\frac{1-x}{1+x}}+\ln \left|\frac{\sqrt{1-x}-\sqrt{1+x}}{\sqrt{1-x}+\sqrt{1+x}}\right|+C$

例题5（较难）：$\int \sqrt{e^{2x}+4e^x-1}dx$ 令$t=e^x,dx=\frac{1}{t}dt$ 即为 $\int \frac{\sqrt{t^2+4t-1}}{t}dt$

解法一：

$\begin{array}{l}=\sqrt{t^2+4t-1}-\int \frac{t^2+2t-t^2-4t+1}{t^2\sqrt{t^2+4t-1}}dt\\ \\ =\sqrt{t^2+4t-1}+\int \frac{2t-1}{t^2\sqrt{t^2+4t-1}}dt\\ \\ =\sqrt{t^2+4t-1}+\int \frac{2}{t\sqrt{t^2+4t-1}}dt-\int \frac{1}{t^2\sqrt{t^2+4t-1}}dt=\sqrt{t^2+4t-1}+4\arctan a-\left(4\arctan a+\frac{4a-2}{1+a^2}\right)\\ \\ =\sqrt{t^2+4t-1}-\frac{2}{t}+\frac{10}{t^2+2t+t\sqrt{t^2+4t-1}}+C\end{array}$

设 $\sqrt{t^2+4t-1}=a-t\to t=\frac{a^2+1}{2a+4},a-t=\frac{a^2+4a-1}{2a+4},dt=\frac{2a^2+8a-2}{{\left(2a+4\right)}^2}da$ （这个方法需要掌握）

$\begin{array}{l}\int \frac{2}{t\sqrt{t^2+4t-1}}dt=2\int \frac{1}{\frac{a^2+1}{2a+4}\cdot \frac{a^2+4a-1}{2a+4}}\cdot \frac{2a^2+8a-2}{{\left(2a+4\right)}^2}da=4\int \frac{1}{a^2+1}da=4\arctan a\\ \\ \int \frac{1}{t^2\sqrt{t^2+4t-1}}dt=\int \frac{1}{{\left(\frac{a^2+1}{2a+4}\right)}^2\cdot \frac{a^2+4a-1}{2a+4}}\cdot \frac{2a^2+8a-2}{{\left(2a+4\right)}^2}da=4\int \frac{a+2}{{\left(a^2+1\right)}^2}da\\ \\ =4\int \frac{a}{{\left(a^2+1\right)}^2}da+8\int \frac{1}{{\left(a^2+1\right)}^2}da=4\arctan a+\frac{4a}{1+a^2}-\frac{2}{a^2+1}=4\arctan a+\frac{4a-2}{1+a^2}\\ \\ 4\int \frac{a}{{\left(a^2+1\right)}^2}da=2\int \frac{d\left(a^2+1\right)}{{\left(a^2+1\right)}^2}=-\frac{2}{a^2+1}\\ \\ 8\int \frac{1}{{\left(a^2+1\right)}^2}da8\int \frac{1}{{\sec }^4\theta }\cdot {\sec }^2\theta d\theta =2\int 1+\cos 2\theta d2\theta =4\theta +2\sin 2\theta =4\arctan a+\frac{4a}{1+a^2}\end{array}$

解法二：(倒代换) 令 $x=\frac{1}{t},dt=-\frac{1}{x^2}dx$

$\begin{array}{l}=\int x\cdot \sqrt{\frac{1+4x-x^2}{x^2}}d\frac{1}{x}=\sqrt{t^2+4t-1}+\int \frac{x-2}{x\sqrt{1+4x-x^2}}dx\\ \\ =\sqrt{t^2+4t-1}+\int \frac{1}{\sqrt{1+4x-x^2}}dx+2\int \frac{-1}{x\sqrt{1+4x-x^2}}dx\\ \\ =\sqrt{t^2+4t-1}+\int \frac{1}{\sqrt{1+4x-x^2}}dx+2\int \frac{\frac{-1}{x^2}}{\sqrt{\frac{1}{x^2}+\frac{4}{x}-1}}dx\\ \\ =\sqrt{t^2+4t-1}+2\ln \left|t+2+\sqrt{t^2+4t-1}\right|+\arcsin \left(\frac{1-2t}{\sqrt{5}t}\right)+C\end{array}$

或 $=\sqrt{t^2+4t-1}+2\ln \left|t+2+\sqrt{t^2+4t-1}\right|-\arctan \left(\frac{2t-1}{\sqrt{t^2+4t-1}}\right)+C$

其中：$\int \frac{1}{\sqrt{1+4x-x^2}}dx=\int \frac{1}{\sqrt{5-{\left(x-2\right)}^2}}dx=\arcsin \left(\frac{x-2}{\sqrt{5}}\right)=\arcsin \left(\frac{1-2t}{\sqrt{5}t}\right)$

$\begin{array}{l}\int \frac{\frac{-1}{x^2}}{\sqrt{\frac{1}{x^2}+\frac{4}{x}-1}}dx=\int \frac{1}{\sqrt{{\left(\frac{1}{x}+2\right)}^2-5}}d\frac{1}{x}\\ \\ =\int \frac{dt}{\sqrt{{\left(t+2\right)}^2-5}}\end{array}$

令 $t+2=\sqrt{5}\sec x,dt=\sqrt{5}\sec x\tan xdx$

$\begin{array}{l}\int \sec xdx=ln\left|\sec x+\tan x\right|=\ln \left|t+2+\sqrt{t^2+4t-1}\right|\end{array}$

练习题1：$\int \frac{\sqrt{1+x+x^2}}{x}$

3. 定积分

3.1 积分定义解题

例1：（定义法求积分）${\int }_a^b\frac{1}{x}dx$

解：令 $x_i=a{\left(\frac{b}{a}\right)}^{\frac{i}{n}},\Delta x=x_{i+1}-x_i=a{\left(\frac{b}{a}\right)}^{\frac{i}{n}}\left[{\left(\frac{b}{a}\right)}^{\frac{1}{n}}-1\right]$

$\begin{array}{l}=\underset{n\to \infty }{\lim }\sum _{i=0}^n\frac{1}{a{\left(\frac{b}{a}\right)}^{\frac{i}{n}}}\cdot a{\left(\frac{b}{a}\right)}^{\frac{i}{n}}\left[{\left(\frac{b}{a}\right)}^{\frac{1}{n}}-1\right]\\ \\ =\underset{n\to \infty }{\lim }n\cdot \left[{\left(\frac{b}{a}\right)}^{\frac{1}{n}}-1\right]=\underset{n\to \infty }{\lim }\frac{\left[{\left(\frac{b}{a}\right)}^{\frac{1}{n}}-1\right]}{\frac{1}{n}}\left(a^x-1\sim x\ln a\right)\\ \\ =\underset{n\to \infty }{\lim }\frac{\frac{1}{n}\ln \left(\frac{b}{a}\right)}{\frac{1}{n}}=\ln \left(\frac{b}{a}\right)\end{array}$

例2：（利用积分定义求极限）

$\begin{array}{l}\mathrm{I}=\underset{n\to \infty }{\lim }\left(\frac{\sin \frac{\pi }{n}}{n+1}+\frac{\sin \frac{2\pi }{n}}{n+\frac{1}{2}}+\cdots +\frac{\sin \frac{n\pi }{n}}{n+\frac{1}{n}}\right)\\ \\ =\underset{n\to \infty }{\lim }\sum _{i=1}^n\frac{\sin \frac{i\pi }{n}}{n+\frac{1}{i}}\ne f\left(\frac{i}{n}\right)\cdot \frac{1}{n}\\ \\ \frac{\sin \pi \frac{i}{n}}{n+1}<\frac{\sin \frac{i\pi }{n}}{n+\frac{1}{i}}<\frac{\sin \pi \frac{i}{n}}{n}\left(i=1,2,\cdots ,n\right)\\ \\ \therefore \underset{n\to \infty }{\lim }\sum _{i=1}^n\sin \pi \frac{i}{n}\cdot \frac{1}{n}={\int }_0^1\sin \pi xdx=-\frac{1}{\pi }\cos \pi x{|}_0^1=\frac{2}{\pi }\\ \\ \underset{n\to \infty }{\lim }\frac{n}{n+1}\cdot \sum _{i=1}^n\sin \pi \frac{i}{n}\cdot \frac{1}{n}=\underset{n\to \infty }{\lim }\frac{n}{n+1}\cdot \underset{n\to \infty }{\lim }\sum _{i=1}^n\sin \pi \frac{i}{n}\cdot \frac{1}{n}\\ \\ =1\cdot {\int }_0^1\sin \pi xdx=\frac{2}{\pi }\end{array}$

有夹逼定理，得 $\mathrm{I}=\underset{n\to \infty }{\lim }\sum _{i=1}^n\frac{\sin \frac{i\pi }{n}}{n+\frac{1}{i}}=\frac{2}{\pi }$

练习题2：

$\begin{array}{l}\left(1\right).\underset{\mathrm{n}\to \infty }{\lim }\ln \sqrt[n]{{\left(1+\frac{1}{n}\right)}^2{\left(1+\frac{2}{n}\right)}^2\cdots {\left(1+\frac{n}{n}\right)}^2}\\ \left(2\right).\underset{n\to \infty }{\lim }\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2^2}}+\cdots +\frac{1}{\sqrt{n^2+n^2}}\right)\end{array}$

3.2 积分技巧解题

例题1：${\int }_0^{\frac{\pi }{4}}\ln (1+tanx)dx$

解：

$\begin{array}{l}={\int }_0^{\frac{\pi }{4}}\ln (1+tan(\frac{\pi }{4}-x))dx\\ \\ ={\int }_0^{\frac{\pi }{4}}\ln (\frac{2}{1+\tan x})dx\\ \\ ={\int }_0^{\frac{\pi }{4}}\ln 2dx-{\int }_0^{\frac{\pi }{4}}\ln (1+tanx)dx\to 原式=\frac{1}{2}{\int }_0^{\frac{\pi }{4}}\ln 2dx=\frac{\pi }{8}\ln 2\end{array}$

例题2：$I={\int }_2^4\frac{\sqrt{x+3}}{\sqrt{x+3}+\sqrt{9-x}}dx$

令 $x+3=9-t,dx=-dt$

$\begin{array}{l}I={\int }_2^4\frac{\sqrt{x+3}}{\sqrt{x+3}+\sqrt{9-x}}dx=-{\int }_4^2\frac{\sqrt{9-t}}{\sqrt{9-t}+\sqrt{3+t}}dt={\int }_2^4\frac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{3+x}}dx\\ \\ =\frac{1}{2}\left[{\int }_2^4\frac{\sqrt{x+3}}{\sqrt{x+3}+\sqrt{9-x}}dx+{\int }_2^4\frac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{3+x}}dx\right]=\frac{1}{2}{\int }_2^{4}dx=1\end{array}$

例题3：${\int }_0^{\mathrm{a}}\sqrt{\frac{a-x}{x+a}}dx$

解：令 $\mathrm{x}=a\cos t,t\in (0,\frac{\pi }{2}),dx=-a\sin tdt$ $\sqrt{\frac{a-x}{x+a}}=\sqrt{\frac{1-\cos t}{1+\cos t}}=\sqrt{\frac{{\sin }^2\frac{t}{2}}{{\cos }^2\frac{t}{2}}}=\tan \frac{t}{2}=\frac{\sin t}{1+\cos t}=\frac{1-\cos t}{\sin t}$ (必须会)

$\begin{array}{l}={\int }_{\frac{\pi }{2}}^0\tan \frac{t}{2}(-asint)dt\\ \\ =a{\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}t}{1+\cos t}dt\\ \\ =a{\int }_0^{\frac{\pi }{2}}\frac{1-{\cos }^{2}t}{1+\cos t}dt\\ \\ =a{\int }_0^{\frac{\pi }{2}}1+\cos tdt\\ \\ =\frac{\pi }{2}a+a\end{array}$

反常积分：

${\int }_0^{\infty }\frac{1}{1+x^4}dx$ 令 $t=\frac{1}{x}$

$$\begin{gathered} {\int }_0^{\infty }\frac{t^2}{1+t^4}dt={\int }_0^{\infty }\frac{x^2}{1+x^4}dx=\frac{1}{2}{\int }_0^{\infty }\frac{1+x^2}{1+x^4}dx \\ =\frac{1}{2}{\int }_0^{\infty }\frac{d\left(x-\frac{1}{x}\right)}{{\left(x-\frac{1}{x}\right)}^2+2}={\frac{1}{2\sqrt{2}}\arctan \frac{1}{\sqrt{2}}\left(x-\frac{1}{x}\right)|}_0^{\infty }=\frac{\pi }{4\sqrt{2}}-\left(-\frac{\pi }{4\sqrt{2}}\right)=\frac{\pi }{2\sqrt{2}} \end{gathered}$$

（记忆）

$\begin{array}{l}{I}_n={\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx={\int }_0^{\frac{\pi }{2}}{\cos }^{n}xdx={\int }_0^{\frac{\pi }{2}}{\sin }^{n-1}xd\left(-\cos x\right)\\ \\ ={\left[-{\sin }^{n-1}x\cos x\right]}_0^{\frac{\pi }{2}}+{\int }_0^{\frac{\pi }{2}}\cos xd\left({\sin }^{n-1}x\right)\\ =\left(n-1\right){\int }_0^{\frac{\pi }{2}}{\sin }^{n-2}x{\cos }^{2}xdx=\left(n-1\right){\int }_0^{\frac{\pi }{2}}{\sin }^{n-2}x\left(1-{\sin }^{2}x\right)dx\\ \\ =\left(n-1\right){\int }_0^{\frac{\pi }{2}}{\sin }^{n-2}xdx-\left(n-1\right){\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx\to I_n=\left(n-1\right)I_{n-2}-\left(n-1\right)I_n\end{array}$

$I_n=\frac{n-1}{n}{I}_{n-2}\to n为奇数I_{2m+1}=\frac{2m\left(2m-2\right)\cdots 42}{\left(2m+1\right)\left(2m-1\right)\cdots 31}=\frac{\left(2m\right)!!}{\left(2m+1\right)!!}$

$n为偶数I_{2m}=\frac{\left(2m-1\right)\left(2m-3\right)\cdots 31}{2m\left(2m-2\right)\cdots 42}=\frac{\left(2m-1\right)!!}{\left(2m\right)!!}\frac{\pi }{2}$

练习题3：

（1）${\int }_0^a{x}^2\sqrt{\frac{a-x}{a+x}}dx$                (2)${\int }_0^1\frac{\ln \left(1+x\right)}{1+x^2}dx$

4. 练习解答

提示 题目解析都是由本人自己所写希望会有不同的做法与我分享，这里就不把答案直接放出来，本人会将答案放在另一篇blogs，等大家做完了点击这里一个链接：答案 。