8.7 不同角度的松树和科松的集成产品

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  将不同角度的松树和科松产品集成在一起

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  The previous concepts looked at integrals where products of same angle trig functions were used. What if the trig function factors have different arguments (angles)? This is the case in many science and engineering problems. One mathematical resource that is extremely useful for modeling some functions is the fourier series, which is an infinite sum of sine waves. In simple terms, a function $\begin{array}{r}S(x)\end{array}$  could be modeled as:
  ::先前的概念查看了使用相同角度三角函数产品的集成体。 如果三角函数系数有不同的论点(矩形)怎么办? 许多科学和工程问题都是如此。对于某些函数建模极为有用的一个数学资源是四倍序列,这是一个无限量的正弦波。简单地说,函数S(x)可以建模为:

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  $$\begin{array}{r}S(x)=a_1\sin x+a_2\sin 2x+a_3\sin 3x\cdots =\underset{n=1}{\overset{\mathrm{\infty }}{\sum }}{a}_n\sin nx.\end{array}$$

  
  ::S(x) = a1sinx+a2sin2x+a3sin3xn=1ansinnx。

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  The weights $\begin{array}{r}{a}_n\end{array}$  reflect which frequencies are most prominent in the modeled function $\begin{array}{r}S(x)\end{array}$ . One of the key characteristics of the Fourier series is the “orthogonality” of each of the sine functions, defined as $\begin{array}{r}\underset{0}{\overset{\pi }{\int }}\sin kx\sin mxdx=0\end{array}$  for $\begin{array}{r}k\ne m\end{array}$ . Can you evaluate the integral $\begin{array}{r}\underset{0}{\overset{\pi }{\int }}\sin kx\sin mxdx=0\end{array}$  to show orthogonality?
  ::在模型函数 S(x) 中最突出的频率的重数。 Fourier 序列的关键特征之一是每个正弦函数的“正弦值”的“正弦值”之一, 定义为 k@ 0sinkxsinmxdx=0。 您能否用 {0sinkxsinmxdx=0 来评价一个元件 {0} 以显示正弦值 ?

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  Integrating Products of Sines and Cosines
  ::Sines和Casines集成产品

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  This section looks at integrals involving the product(s) of sine and cosine functions having different linear arguments, e.g. $\begin{array}{r}\int \sin (ax)\sin (bx)dx\end{array}$ , or $\begin{array}{r}\int {\sin }^2(ax)\cos (cx)dx\end{array}$ . Use is made of the following three basic sine and cosine product identities involving different arguments:
  ::本节探讨具有不同线性参数的正弦和共弦函数产品和共弦函数的集成,如sin(ax)sin(bx)dx,或sin2(ax)cos(cx)dx。 使用下列三个基本正弦和共弦产品特性,涉及不同参数:

  播放段落 Sine/Cosine Product Identity ::松弦/焦豆产品标识	播放段落 Derived From ::调自
  播放段落 $\begin{array}{r}\sin A\sin B=\frac{1}{2}\left[\cos (A-B)-\cos (A+B)\right]\end{array}$ ::AsinB=12 [cos(A-B)-cos(A+B)]	播放段落 $\begin{array}{r}\cos (A+B)=\cos A\cos B-\sin A\sin B\end{array}$ ::cos(A+B) =cosAcosB-sinAsinB 播放段落 $\begin{array}{r}\cos (A-B)=\cos A\cos B+\sin A\sin B\end{array}$ ::cos(A-B) =cosAcosB+sinAsinB
  播放段落 $\begin{array}{r}\sin A\cos B=\frac{1}{2}\left[\sin (A-B)+\sin (A+B)\right]\end{array}$ ::AcosB=12[sin(A-B)+sin(A+B)]	播放段落 $\begin{array}{r}\sin (A+B)=\sin A\cos B-\cos A\sin B\end{array}$ :A+B)=sinAcosB-cosAsinB 播放段落 $\begin{array}{r}\sin (A-B)=\sin A\cos B+\cos A\sin B\end{array}$ :A-B)=sinAcosB+cosAsinB
  播放段落 $\begin{array}{r}\cos A\cos B=\frac{1}{2}\left[\cos (A-B)+\cos (A+B)\right]\end{array}$ ::[Cos(A-B)+Cos(A+B)]	播放段落 $\begin{array}{r}\cos (A+B)=\cos A\cos B-\sin A\sin B\end{array}$ ::cos(A+B) =cosAcosB-sinAsinB 播放段落 $\begin{array}{r}\cos (A-B)=\cos A\cos B+\sin A\sin B\end{array}$ ::cos(A-B) =cosAcosB+sinAsinB

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  The procedure involves applying the identities to reduce the complexity of the integrand .
  ::该程序涉及应用身份来降低原数的复杂性。

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  For example, evaluate  $\begin{array}{r}\int \sin x\sin 3xdx\end{array}$ .
  ::例如,评估 sinxsin3xdx。

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  $$\begin{array}{rl}\int \sin x\sin 3xdx& =\int \frac{1}{2}\left[\cos (x-3x)-\cos (x+3x)\right]dx\\ & =\int \frac{1}{2}\left[\cos (-2x)-\cos (4x)\right]dx\\ & =\frac{1}{2}\left[\frac{\sin (-2x)}{-2}\right]-\frac{1}{2}\left[\frac{\sin 4x}{4}\right]\\ \int \sin x\sin 3xdx& =\frac{\sin (2x)}{4}-\frac{\sin 4x}{8}+C\end{array}$$

  
  :x+3x)]

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to evaluate the integral $\begin{array}{r}\underset{0}{\overset{\pi }{\int }}\sin kx\sin mxdx=0\end{array}$  to show orthogonality.
  ::先前,您被要求评估 =0sinkxsinmxdx=0 以显示正向性 。

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  Option 1: Using integration by parts (2 passes required) to solve for the integral yields
  ::备选办法1:使用各部件集成的办法(2张所需通行证)解决整体产量问题

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  $$\begin{array}{r}\underset{0}{\overset{\pi }{\int }}\sin kx\sin mxdx=\frac{1}{m^2-k^2}[k\cos k\pi \sin m\pi -m\sin k\pi \cos m\pi ].\end{array}$$

  
  ::@ksinkxsinmxdx=1m2-k2 [kcosksinmmsinkkcosm]。

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  For the integral to be 0,  $\begin{array}{r}m\end{array}$ and  $\begin{array}{r}k\end{array}$ must be integers $\begin{array}{r}(\sin m\pi =\sin k\pi =0)\end{array}$
  ::组合体为 0, m 和 k 必须为整数( sinmmsink0)

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  Option 2: Using sum and difference angle properties for the integrand yields $\begin{array}{r}\sin kx\sin mx=\frac{1}{2}[\cos (k-m)x-\cos (k+m)x]\end{array}$ . The integral becomes
  ::选项2:使用总和和差角属性来表示整数 sinçkxsinmx=12 [cos(k-m)x-cos(k+m)x]。

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  $$\begin{array}{r}\underset{0}{\overset{\pi }{\int }}\sin kx\sin mxdx=\frac{1}{2(k^2-m^2)}[(k+m)\sin (k-m)\pi -(k-m)\sin (k+m)\pi ].\end{array}$$

  
  ::=============================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================

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  For the integral to be 0, $\begin{array}{r}(k-m)\end{array}$  and $\begin{array}{r}(k+m)\end{array}$  must be integers.
  ::整体值为 0, (k-m) 和 (k+m) 必须为整数。

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  Example 2
  ::例2

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  Evaluate $\begin{array}{r}\int \sin (9x)\cos (4x)dx\end{array}$
  ::评估 sin ( 9x) cos ( 4x) dx

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  $$\begin{array}{rl}\int \sin (9x)\cos (4x)dx& =\int \frac{1}{2}\left[\sin (9x-4x)+\sin (9x+4x)\right]dx\\ & =\frac{1}{2}\int \left[\sin (5x)+\sin (13x)\right]dx\\ & =\frac{1}{2}\left[\frac{-\cos (5x)}{5}\right]+\frac{1}{2}\left[\frac{-\cos 13x}{13}\right]\\ \int \sin (9x)\cos (4x)dx& =-\left[\frac{\cos (5x)}{10}+\frac{\cos (13x)}{26}\right]+C\end{array}$$

  
  : 9x) ( 9x) ( 4x) dx ( 9x) ( 9x4x) +sin ( 9x+4x) dx= 12 <sin( 5x) +sin( 13x) dx= 12 [ -cos( 5x) ( 5x) 5] +12 [ -cos( 5x) ( 13x13) +12 [ -cos ( 9x) ( 4x) dx [cos( 5x) 10+cos( 13x) 26]+C

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  Example 3
  ::例3

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  Evaluate $\begin{array}{r}\int \cos (9x)\cos (5x)dx\end{array}$
  ::评估 (9x) (5x) dx

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  $$\begin{array}{rl}\int \cos (9x)\cos (5x)dx& =\int \frac{1}{2}\left[\cos (9x-4x)+\cos (9x+4x)\right]dx\\ & =\frac{1}{2}\int \left[\cos (5x)+\cos (13x)\right]dx\\ & =\frac{1}{2}\left[\frac{\sin (5x)}{5}\right]+\frac{1}{2}\left[\frac{\sin 13x}{13}\right]+C\\ \int \sin (9x)\cos (4x)dx& =\left[\frac{\sin (5x)}{10}+\frac{\sin 13x}{26}\right]+C\end{array}$$

  
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  Example 4
  ::例4

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  Evaluate the integral $\begin{array}{r}\int {\sin }^{2}2x\cos 7xdx\end{array}$
  ::评估集成 @sin2 @%2xcos=7xdx

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  $$\begin{array}{rl}\int {\sin }^{2}2x\cos 7xdx& =\int \frac{1}{2}(1-\cos 4x)\cos 7xdx\\ & =\int \frac{1}{2}(\cos 7x-\cos 4x\cos 7x)xdx\\ & =\frac{1}{2}\int \cos 7xdx-\frac{1}{2}\int \cos 4x\cos 7xdx\\ & =\frac{1}{14}\sin 7x-\frac{1}{2}\int \frac{1}{2}[\cos 11x+\cos 4x]dx\\ & =\frac{1}{14}\sin 7x-\frac{1}{4}\left[\frac{\sin 11x}{11}+\frac{\sin 4x}{4}\right]+C\\ & =\frac{1}{14}\sin 7x-\frac{1}{44}\sin 11x-\frac{1}{12}\sin 3x+C\end{array}$$

  
  ::2222xxcos 7xxxl12(1-cos4xx)cos 7xxx12xxxx=127xxxxx=12cos7xxxxxx*127xxxxxxxxx=114sin7xx4xdxx=114sin7xxx14[sin11x11+sin4x44]+C=114sin7xx-144sin*11x-11xxxx11xxxx1x112xxxx14}[cos11x+csusxxxxxx}3x+C

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  Review
  ::回顾

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  Evaluate the integrals:
  ::评估整体体:

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  1. $\begin{array}{r}\int \sin 2x\cos 3xdx\end{array}$
     ::2xcos
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  2. $\begin{array}{r}\int \sin 4x\sin 3xdx\end{array}$
     ::3xdx
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  3. $\begin{array}{r}\int \cos 3x\cos 4xdx\end{array}$
     ::============================================================================================================================================= ==============================================================================================================================
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  4. $\begin{array}{r}\int \cos 2x\cos 3xdx\end{array}$
     ::========================================================================================================================================
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  5. $\begin{array}{r}\int \cos 2x\sin 3xdx\end{array}$
     ::============================================================================================================================================== ============================================================================================
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  6. $\begin{array}{r}\int \sin 2x\sin 3xdx\end{array}$
     ::2xsin3xxxxx
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  7. $\begin{array}{r}\underset{0}{\overset{\frac{\pi }{2}}{\int }}4\sin x\cos (\frac{x}{3})dx\end{array}$
     ::024sinxcos( x3) dx
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  8. $\begin{array}{r}\underset{0}{\overset{\frac{\pi }{2}}{\int }}4\sin (\frac{x}{3})\cos (x)dx\end{array}$
     ::024sin(x3)cos(x)dx
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  9. $\begin{array}{r}\int \sin (6x)\sin (2x)dx\end{array}$
     :6x) (2x) dx
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  10. $\begin{array}{r}\int \cos (2x)\cos (-x)dx\end{array}$
      :2x) (-x) dx
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  11. $\begin{array}{r}\int \sin (10x)\cos (5x)dx\end{array}$
      :10x) (5x) dx
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  12. $\begin{array}{r}\int {\sin }^3(3x)\cos (5x)dx\end{array}$
      :3x) (5x) dx
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  13. $\begin{array}{r}\underset{0}{\overset{2}{\int }}\cos (3\pi x){\cos }^2(\pi x)dx\end{array}$
      ::*02cos*(3xx)cos2*(xx)dx
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  14. $\begin{array}{r}\int x\sin 2x\sin 3xdx\end{array}$
      ::
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  15. $\begin{array}{r}\int x\cos 2x\cos 3xdx\end{array}$
      ::

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  Review (Answers)
  ::回顾(答复)

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  Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
  ::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。