2025年第四单元用python学习微积分（二十五）三角函数，反向变量替换，配方，三角函数变换表

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本文内容来自于学习麻省理工学院公开课：单变量微积分-反向变量替换，配方-网易公开课

开发环境准备：CSDN

一、公式梳理

1、 (secant)

2、 (cosecant)

3、 (tangent)

4、 (cotangent)

5、

6、

7、

8、

9、

二、例题

1、

2、

3、如上题去除表达式 " tan(arccsc(x)) " 中三角函数

三、三角替换的小结

1、表

2、例题

一、公式梳理

1、 $sec(\theta) = \frac{1}{cos(\theta)}$
讯享网 (secant)

2、 $cec(\theta) = \frac{1}{sin(\theta)}$ (cosecant)

3、 $tan(\theta) = \frac{sin(\theta)}{cos(\theta)}$ (tangent)

4、 $cot(\theta) = \frac{cos(\theta)}{sin(\theta)}$ (cotangent)

5、 $sec^2(\theta) = \frac{cos^2(\theta)+sin^2(\theta)}{cos^2(\theta)} =1+tan^2(\theta)$

6、 $tan'(\theta) = (\frac{sin(\theta)}{cos(\theta)})' = \frac{cos^2(\theta)+sin^2(\theta)}{cos^2(\theta)} = \frac{1} {cos^2(\theta)} = sec^2(\theta)$

7、 $sec'(\theta) = (\frac{1}{cos(\theta)})' = \frac{-(-sin(\theta))} {cos^2(\theta)} = tan(\theta)sec(\theta)$

8、 $\int tan(x)dx = \int \frac{sin(x)}{cos(x)}dx$

设 u = cos(x), u' =du = -sin(x)dx

$\int \frac{-du}{u} = -ln(u) +c = -ln(cos(x)) +c$

9、 $\int sec(x) dx$

设 u = tan(x) + sec(x)

由公式 6, 7

du = sec^2(x) + tan(x)sec(x) = sec(x)(sec(x) + tan(x))dx

$sec(x)dx = \frac{du}{u}$

$\int sec(x) dx = \int \frac{du}{u} = ln(u) +c = ln(tan(x) + sec(x) ) +c$

二、例题

1、 $\int sec^4(x)dx = \int (1+tan^2(x))sec^2(x)dx$

u = tan(x) ; du = sec^2(x)dx

$\int (1+u^2)du = u +\frac{u^3}{3} +c= tan(x) + \frac{tan^3(x)}{3}+c$

用定积分做个简单的验算，取值 $\in (0 , \frac{\pi}{3})$ ，结果是符合的

import numpy as np from sympy import * import matplotlib.pyplot as plt def DrawXY(xFrom,xTo,steps,expr,color,label,plt): yarr = [] xarr = np.linspace(xFrom ,xTo, steps) for xval in xarr: yval = expr.subs(x,xval) yarr.append(yval) y_nparr = np.array(yarr) plt.plot(xarr, y_nparr, c=color, label=label) def DrawTrapezoidal(xFrom,xTo,steps,expr,color,plt, label=''): width = (xTo - xFrom)/steps xarrRect = [] yarrRect = [] area = 0 xprev = xFrom yvalAll = 0 for step in range(steps): yval1 =expr.subs(x,xprev) yval2 = expr.subs(x,xprev + width) xarrRect.append(xprev) xarrRect.append(xprev) xarrRect.append(xprev + width) xarrRect.append(xprev + width) xarrRect.append(xprev) yarrRect.append(0) yarrRect.append(yval1) yarrRect.append(yval2) yarrRect.append(0) yarrRect.append(0) area += width * (yval1+yval2)/2 plt.plot(xarrRect, yarrRect, c=color) xprev= xprev + width print('============================') if len(label)!=0: print(label) print('============================') print('width = ', width) print('area = ',area) x = symbols('x') expr = sec(x)4 DrawXY (0,np.pi/3,1000, expr, 'c', label='sec(x)',plt=plt) print(integrate(expr,(x,0,np.pi/3))) print(tan(np.pi/3)+tan(np.pi/3)3/3 - tan(0)+tan(0)3/3) plt.legend(loc='upper left') plt.show()

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添加图片注释，不超过 140 字（可选）

2、 $\int \frac{dx}{x^2\sqrt{1+x^2}}$

由于tan取值范围从负无穷一直到正无穷，所以可以进行三角替换

设 $tan(\theta) = x 则 dx = tan'(\theta) = sec^2(\theta)d\theta$

另由 $1+tan^2(\theta) =sec^2$

$\int \frac{dx}{x^2\sqrt{1+x^2}} = \int \frac{sec^2(\theta)d\theta}{tan^2(\theta)\sqrt{1+tan^2(\theta)}}$

$= \int \frac{sec^2(\theta)d\theta}{tan^2(\theta)\sqrt{sec^2(\theta)}} = \int \frac{sec(\theta) d\theta}{tan^2(\theta)} = \int \frac{\frac{1}{cos(\theta)}d\theta}{\frac{sin^2(\theta)}{cos^2(\theta)}}$

$= \int \frac{cos(\theta)}{sin^2(\theta)} d\theta$

设 $u = sin(\theta) , du = cos(\theta)d\theta$

$= \int \frac{du}{u^2} = \frac{1}{-u} +c= \frac{1}{-sin(\theta)}+c = -csc(\theta) +c$

还原三角替换

在上图中直角三角形中，设 $\theta$ 的对边长为x，则根据前面假设tan(\theta) = x , 有临边长为1，另斜边长则为 $\sqrt{1+x^2}$

$-csc(\theta) +c = -\frac{1}{sin(\theta)} + c =-\frac{\sqrt{1+x^2}}{x} +c$

或由 $tan(\theta) = x$

$-csc(\theta) +c = -csc( arctan(x)) +c$

用定积分验算下，用梯形法得到的结果和计算出的结果近似。

讯享网x = symbols('x') expr = 1/ (x2*(1+x2)0.5) DrawXY (2,5,100, expr, 'c', label='',plt=plt) DrawTrapezoidal(2,5,100,expr,'c',plt, label='1/ (x2*(1+x2)0.5)') #print(integrate(expr,(x,2,5))) print('result with arc tangent=', float(-csc(atan(5)) + csc(atan(2)))) print('result = ', -((1+52)0.5)/5 + ((1+22)0.5)/2 )

3、如上题去除表达式 " tan(arccsc(x)) " 中三角函数

theta = 45 angle = np.linspace( 0 , 2 * np.pi , 150 ) radius = 1 xarr = radius * np.cos( angle ) yarr = radius * np.sin( angle ) figure, axes = plt.subplots( 1 ) axes.plot( xarr, yarr, label='radius='+format(radius) ) angleInner = np.linspace( theta/180*np.pi , 0 , 150 ) xArcInner = radius /10 * np.cos( angleInner ) yArcInner = radius /10 * np.sin( angleInner ) axes.plot( xArcInner, yArcInner,color='b') plt.text(0.18, 0.03, 'θ', fontsize=9) plt.text(cos(theta)-0.18, -0.08, 'adj', fontsize=9) plt.text(cos(theta)+0.22, sin(theta)-0.48, 'opp', fontsize=9) plt.text(cos(theta)-0.3, sin(theta)-0.48, 'hyp', fontsize=9) axes.set_aspect( 1 ) x1 = np.cos( theta/180*np.pi ) y1 = np.sin( theta/180*np.pi ) xarr1 = [0,1] yarr1 = [0,0] plt.plot(xarr1, yarr1,'r',linestyle='-',marker='') xarr2 = [0,x1] yarr2 = [0,y1] plt.plot(xarr2, yarr2,'orange',linestyle='-',marker='') plt.plot([x1,x1], [0,y1],'green',linestyle='-',marker='') plt.plot([0,x1], [0,0],'green',linestyle='-',marker='') plt.text(x1+0.05, y1+0.05, '(cosθ ,sinθ)', fontsize=9) plt.text(-0.1, -0.1, 'O', fontsize=9) plt.legend(loc='upper left') plt.show()

$csc(\theta) = x$

$\frac{1}{sin(\theta)} =x = \frac{hyp}{opp}$

设斜边(hyp)为x，则对边(opp)为1, 临边(adj)为 $\sqrt{x^2 -1}$

$tan(arccsc(x)) = tan(\theta) = \frac{opp}{adj} = \frac{1}{\sqrt{x^2 -1}}$

三、三角替换的小结

1、表

如果积分包含 -- 要做的替换 -- 得到

$\sqrt{a^2 -x^2}$ -- $x = a \times cos(\theta)$ or $x = a \times sin(\theta)$ -- $a \times sin(\theta)$ or $a \times cos(\theta)$

当 $x = a \times cos(\theta)$ ， $dx = -a*sin(\theta)d\theta$

当 $x = a \times sin(\theta)$ ， $dx = a*cos(\theta)d\theta$

$\sqrt{a^2 +x^2}$ -- $x = a \times tan(\theta)$ -- $a \times sec(\theta)$ -- $dx = asec(\theta)^2d\theta$

$\sqrt{x^2 -a^2}$ -- $x = a \times sec(\theta)$ -- $a \times tan(\theta)$ -- $dx = atan(\theta)sec(\theta)d\theta$

$\sqrt{a^2 + x^2} = \sqrt{a^2 + a^2tan(x)} = a \times \sqrt{1+tan^2(x)}=a\sqrt{\frac{cos^2(x)}{cos^2(x)}+\frac{sin^2(x)}{cos^2(x)}} =a\sqrt{\frac{1}{cos^2(x)}} =a \times sec(x)$