定积分七绝🗡

Author:张一极 🔱

个人博客:极度空间

公众号:视觉迷航

T : 原函数

一绝:奇云化雨

遇到可以在原函数中分离出奇函数,且积分区间为0点两侧对称区间

$$\int_{-1}^{1} \frac{x+1}{1+\sqrt[3]{x^{2}}} d x$$

分离求解

$$=\int_{-1}^{1} \frac{x}{1+\sqrt[3]{x^{2}}} \mathrm{d} x+\int_{-1}^{1} \frac{1}{1+\sqrt[3]{x^{2}}} \mathrm{d} x=0+2 \int_{0}^{1} \frac{1}{1+\sqrt[3]{x^{2}}} \mathrm{d}$$

剩下部分,换元以后可解.

二绝:取常救主

积分单一形式复杂,换元求解有望拆解的形式,积分含可提常数经常使用分子分母提取常数,顺利抽取以后化简结果

 

$$2\int_{0}^{1} \frac{1}{1+\sqrt[3]{x^{2}}} \mathrm{d} \\t = \sqrt[3]{x^{2}}\\ \therefore x = \sqrt{t^3}\\dx = \frac{2}{3}t^{\frac{1}{2}}\\ T=3\int_{0}^{1}\frac{1\cdot \sqrt t}{1+t}dt\\$$

已经进一步接近arctanx的导数形式,再进行一轮换元:

$$\because \sqrt t=i\\i=0,t=0\\i=1,t=1\\积分上下限不变\\ \therefore T= 3\cdot 2\int_{0}^{1}{\frac{\sqrt u \cdot \sqrt{u}}{1+u}du}= 6\int_{0}^{1}{\frac{{u}}{1+u}du}\\ 6\int_{0}^{1}{\frac{{u}}{1+u}du} =6\int_{0}^{1}{\frac{u+1}{1+u}}du-6\int_{0}^{1}{\frac{1}{1+u}}du =6-arctanx|^1_0$$

三绝:投桃报李

 

$$\int_{0}^{2 \pi} \frac{1}{1+\cos ^{2} x} \mathrm{d} x$$

要实现三角函数代换,一定要积分区间里实现连续函数可导

利用积分函数奇偶性化简:

$$\int_{0}^{2 \pi} \frac{1}{1+\cos ^{2} x} \mathrm{d} x \\ \because\int_{-\pi}^{\pi}\frac{1}{1+cos^2x}dx=2\int_{0}^{\pi}\frac{1}{1+cos^2x}dx \\T=2(\int_{0}^{\frac{\pi}{2}}\frac{1}{1+cos^2x}dx+\int_{\pi/2}^{\pi}...dx) \\x==\pi-t \\T=4\int_{0}^{\frac{2}{\pi}}\frac{1}{1+cos^2x}dx \\类比arctanx的导数形式,取t=arctanx,根据三角边,斜边为\sqrt{1+t^2},cosx = \frac{1}{\sqrt{1+t^2}}\\T=4\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\frac{1}{1+t^2}}dx\\ T= \int_{0}^{1}\frac{1}{\frac{2+t^2}{1+t^2}}\frac{1}{1+t^2}dt \\=4\int_{0}^{1}\frac{1}{2+t^2}dt\\ 变上下限为|^1_0 \\T= \frac{4}{\sqrt{2}}arctan(\frac{t}{\sqrt{2}})|^1_0$$

 

四绝:偶倍奇零

$$\int_{0}^{2 \pi} \frac{1}{1+\cos ^{2} x} \mathrm{d} x \\ \because\int_{-\pi}^{\pi}\frac{1}{1+cos^2x}dx=2\int_{0}^{\pi}\frac{1}{1+cos^2x}dx \\T=2(\int_{0}^{\frac{\pi}{2}}\frac{1}{1+cos^2x}dx+\int_{\pi/2}^{\pi}...dx)$$

五绝:精准打击

$$\lim _{n \rightarrow \infty}\left(\frac{n+1}{n^{2}+1}+\frac{n+2}{n^{2}+4}+\frac{n+3}{n^{2}+9}+\dots+\frac{n+n}{n^{2}+n^{2}}\right)$$

放缩无效的精准定义解法

1.提取$\frac{1}{n}$作为dx,取另外i为一变量

2.凑出$\frac{i}{n}$

3.取$\frac{1}{n}$为dx,$\frac{i}{n}$为x

$$\frac{n+i}{n^2+i^2}=\frac{1}{n}\cdot\frac{n+i}{n+\frac{i^2}{n}}<br>\\凑\frac{i}{n}<br>\\T=\frac{1}{n}\cdot\frac{1+\frac{i}{n}}{1+\frac{i^2}{n^2}}<br>\\\frac{i}{n}=x\\<br>T= \int_{0}^{1}\frac{1+x}{1+x^2}dx<br>\\取积分得:arctanx|^1_0+\frac{1}{2}ln(1+x^2)|^1_0<br>\\=\frac{\pi}{4}+\frac{1}{2}ln2$$

 

六绝:区间再现

$$\int_{a}^{b}f(x)dx\to x = a+b-t \\\therefore T=\int_{b}^{a}f(a+b-t)-d(t)=\int_{a}^{b}f(a+b-t)dt \\=\int_{a}^{b}f(a+b)dt-\int_{a}^{b}f(t)dt=\int_{a}^{b}f(a+b)dt-\int_{a}^{b}f(x)dx\\T=\frac{1}{2}\int_{a}^{b}f(a+b)dt$$

通过特殊换元,得到一个新的式子,包含原来的式子,区间再现

七绝:n阶迭代

$$\int_{0}^{\frac{\pi}{2}} \sin ^{n} x d x=\left\{\begin{array}{l}\frac{n-1}{n} \cdot \frac{n-3}{n-2} \cdot \cdots \cdot \frac{2}{3} \cdot I_{1}, \\ \frac{n-1}{n} \cdot \frac{n-3}{n-2} \cdot \cdots \cdot \frac{1}{2} \cdot I_{0}\end{array}=\left\{\begin{array}{l}\frac{n-1}{n} \cdot \frac{n-3}{n-2} \cdots \cdots \frac{2}{3} \cdot 1, n \text { 为大于一的奇数 } \\ \frac{n-1}{n} \cdot \frac{n-3}{n-2} \cdots \cdots \frac{1}{2} \cdot \frac{\pi}{2}, n \text { 为正偶数. }\end{array}\right.\right.$$

$$例如:\begin{array}{l}\int_{0}^{\frac{\pi}{2}} \sin ^{8} x \mathrm{d} x=\frac{7}{8} \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} \\ \int_{0}^{\frac{\pi}{2}} \sin ^{9} x \mathrm{d} x=\frac{8}{9} \cdot \frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3} \cdot 1\end{array}$$

$$\int_{0}^{\pi} x \sin ^{9} x d x \\x = \pi-t \\T=\int_{0}^{\pi}(\pi-t)sin^9(\pi-t)dt-根据六绝 \\T=\frac{\pi}{2}\int_{0}^{\pi}sin^9xdx \\=\frac{\pi}{2}\int_{0}^{\frac{\pi}{2}} \sin ^{9} x \mathrm{d} x=\frac{8}{9} \cdot \frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3} \cdot 1$$

推广:

 

$$\int_{0}^{2 \pi} \cos ^{n} x \mathrm{d} x=\int_{0}^{2 \pi} \sin ^{n} x \mathrm{d} x=\left\{\begin{array}{ll}0, & n \text { 为正奇数 } \\ 4 \cdot \frac{n-1}{n} \cdot \frac{n-3}{n-2} \cdots \cdots \cdot \frac{1}{2} \cdot \frac{\pi}{2}, & n \text { 为正偶数. }\end{array}\right.$$