Section: 按部分集成: 解决未知集成 | 8.3 按部分整合:解决未知综合体

Section outline

In the previous concept, you learned that more than one application of the integration by parts technique might be required to solve an integral $\begin{aligned} \int F (x) d x \end{aligned}$ . In this concept, you will encounter integrals that require a multiple pass approach and in the process generate the same integral you are trying to solve. When this happens, a simple algebraic manipulation can be used to solve for the unknown integral. Can you solve for the integral $\begin{aligned} \int \cos 5 x \sin 7 x d x \end{aligned}$ in the second application of integration by parts pass?
::在前一个概念中,您了解到,要解析 F (x) dx 的集成,可能需要不止一个应用部件集成技术。在这个概念中,您会遇到需要多重传承法的集成体,而在此过程中,您正在试图解析的集成体会产生同样的集成体。一旦发生这种情况,可以使用简单的代数操纵来解决未知的集成体。在第二个集成应用中,您能够解析 cos5xsin7xdx 的集成体吗?

Solving for the Unknown Integral
::未知集成器的解决

In the previous concept, we used the technique of integration by parts (see below) multiple times to transform a “difficult” integral to successively “easier” integrals, the last of which could be evaluated to complete the solution.
::在前一个概念中,我们多次使用按部分(见下文)分类的整合技术(见下文),以改变相继“容易”的组合体所固有的“困难”,对最后的组合体可进行评估,以完成解决方案。

Integration by Parts
::按部分分列的整合情况

Evaluate the “difficult” integral $\begin{aligned} \int F (x) d x = \int f (x) g^{'} (x) d x \end{aligned}$ .
::评估“困难”的“困难”组成部分 @F(x)dxf(x)g_(x)dx。

Transform $\begin{aligned} \int f (x) g^{'} (x) d x \end{aligned}$ to the form $\begin{aligned} \int u d v = u v - \int v d u \end{aligned}$ with the right choice for $\begin{aligned} u = f (x) \end{aligned}$ and $\begin{aligned} d v = g^{'} (x) d x \end{aligned}$ that makes $\begin{aligned} \int v d u \end{aligned}$ easier to evaluate than $\begin{aligned} \int u d v \end{aligned}$ .
::f(x)g(x)dx 至窗体 udv=uvvdu, 且对 u=f(x) 和 dv=g=g}(x)dx 进行正确选择, 使得 vdu 的评审比 udv 更容易。

Guide:
::指南:

Choose $\begin{aligned} d v \end{aligned}$ to be the more complicated portion of the integrand that fits a basic integration formula. Choose $\begin{aligned} u \end{aligned}$ to be the remaining term in the integrand.
::选择 dv 是符合基本集成公式的整数中较复杂的部分。选择 u 是整数中的剩余词。

Choose $\begin{aligned} u \end{aligned}$ to be the portion of the integrand whose derivative is simpler than $\begin{aligned} u \end{aligned}$ . Choose $\begin{aligned} d v \end{aligned}$ to be the remaining term.
::选择 u 成为 integrand 的一部分, 其衍生值比 u 简单。选择 dv 作为剩余任期。

Remember, the goal of the integration by parts is to start with an integral in the form $\begin{aligned} \int u d v \end{aligned}$ that is hard to integrate directly and change it to an integral $\begin{aligned} \int v d u \end{aligned}$ that is easier to evaluate. However, there are some integrals $\begin{aligned} \int u d v \end{aligned}$ where multiple iterations of integration by parts yields a version of $\begin{aligned} \int v d u \end{aligned}$ that is the original integral. In these cases a solution is found using simple algebra. These kinds of integrals crop up often in electrical engineering and other disciplines.
::记住,各部分集成的目的是从一个以 udv 形式的有机体开始,这种有机体很难直接集成,并改变成一个易于评估的有机体 vdu。然而,有一些有机体 udv , 部件集成的多重迭代产生一种 vdu 的原始有机体。在这种情况下, 使用简单的代数可以找到一个解决方案。这些有机体往往在电气工程和其他学科中生长。

Take the integral $\begin{aligned} \int e^{x} \cos x d x \end{aligned}$ .
::使用积分 excos xdx 。

$\begin{aligned} Use : \int u d v = u v - \int v d u . \\ Choose : u = \cos x d v = e^{x} d x \\ Form : d u = - \sin x d x v = e^{x} \\ Evaluate : \int e^{x} \cos x d x = e^{x} \cos x + \int e^{x} \sin x d x \dots The new integral looks similar to the old integral! \end{aligned}$

::使用 : @ udv=uvvdu. choose: u=cosxdv=exdxForm: dosinxdxv=exEvaluate:\ excosxdx=excosxxxxxsinxdxxxx}}}# 新的集成与旧的集成类似 !

Notice that the second integral looks the same as our original integral in form, except that it has a $\begin{aligned} \sin x \end{aligned}$ instead of $\begin{aligned} \cos x \end{aligned}$ . To evaluate it, we again apply integration by parts to the second term:
::请注意,第二个整体体看起来与我们原有整体体的形式相同,但有一个sinx而不是cosx。要评估它,我们再次对第二个术语部分地适用合并:

$\begin{aligned} Use : & \int u d v = u v - \int v d u . \\ Choose : & u = \sin x d v = e^{x} d x \\ Form : & d u = \cos x d x v = e^{x} \\ Evaluate : & \int e^{x} \cos x d x = e^{x} \cos x + e^{x} \sin x - \int e^{x} \cos x d x \dots The new integral looks the \\ same as the original integral! \end{aligned}$

::使用 : @ udv=uvvdu. choose: u=sinxdv=exdxForm: du=cosxdxv=exEvaluate:\ excosxdx=excosx+exsinxxexcosxxxdxxx... 新的组合将它们看成是原始的组合!

Notice that the unknown integral now appears on both sides of the equation. We can use simple algebra to move the unknown integral on the right to the left side of the equation to solve for the integral:
::注意未知的元件现在出现在方程式的两侧。我们可以使用简单的代数将方程式右侧的未知元件移到方程式的左侧, 以解决元件 :

$\begin{aligned} 2 \int e^{x} \cos x d x & = e^{x} \cos x + e^{x} \sin x + C \\ \int e^{x} \cos x d x & = \frac{1}{2} [\cos x + \sin x] e^{x} + C \dots Use the same dummy constant of integration . \end{aligned}$

::2 excosxdx=excosx+exsinx+Cçexcosxdx=12 [cosx+sinx]ex+C... 使用相同的模拟整合常数。

A two-pass approach has been used to obtain the solution by getting back to the original integral.
::现已采用双向办法,回到原来的整体体,从而获得解决办法。

Now, take the integral $\begin{aligned} \int \cos (\ln x) d x \end{aligned}$ .
::现在,拿一个完整的(xxx)dx。

$\begin{aligned} Use : & \int u d v = u v - \int v d u . \\ Choose : & u = \cos (\ln x) d v = d x \\ Form : & d u = - \frac{\sin (\ln x)}{x} d x v = x \\ Evaluate : & \int \cos (\ln x) d x = x \cos (\ln x) - \int - x \frac{\sin (\ln x)}{x} d x \\ = x \cos (\ln x) + \int \sin (\ln x) d x \dots The new integral looks similar to the old integral! \end{aligned}$

::使用 : @ udv=uvvdu. choose: u=cos( lnx) dv=dxForm: dusin( lnx) xdxv=xEvaluate:\\ cos( lnx) dx=xcosí( lnx) xdxx=xcosí}( lnx) xdxx=xcos( lnx) {( lnx)\\\\\\\\\\\\\\\\ (lnx) dx\\\\\\\\ xxxx\\\\\\\ 新的内装与旧的内装类似 !

The second integral looks similar to our original integral in form, except that it has a $\begin{aligned} \sin (\ln x) \end{aligned}$ instead of $\begin{aligned} \cos (\ln x) \end{aligned}$ . To evaluate it, we again apply integration by parts:
::第二个整体体看起来与我们原来的整体体相似,只是它有一个罪而不是cos(lnx)。为了评估它,我们再次按部分进行整合:

$\begin{aligned} Use : \int u d v = u v - \int v d u . \\ Choose : u = \sin (\ln x) d v = d x \\ From : d u = \frac{\cos (\ln x)}{x} d x v = x \\ Evaluate : \int \cos (\ln x) d x = x \cos (\ln x) + [x \sin (\ln x) - \int x \frac{\cos (\ln x)}{x} d x] \\ = x \cos (\ln x) + [x \sin (\ln x) - \int \cos (\ln x) d x] \dots The new integral looks \\ the same as the old integral! \end{aligned}$

::使用 : @ udv=uvvdu. Choose: u=sin( lnx) dv= dxfrom: du=cos( lnx) xdxv=xEvaluate: @ cos( lnx) dx=xcos( lnx) xxxxx) +xcos( lnx) +[ xsin( xx) +[ xsin( xxx) {xdxv=xvaluate: @ cos( xx) xdxx=xvaluate:\\\\\\\ xxxxxxxxx=xcos* (lnx) +[ xxxxxxxxxx] =xcos=xcos=xxxx{( =xin( x_xxxxxxxxxxxxxxxxxx) =xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx) !!!!

Again, notice that the unknown integral now appears on both sides of the equation. We can use simple algebra to move the unknown integral on the right to the left side of the equation to solve for the integral:
::请注意,未知的元件现在出现在方程式的两侧。我们可以使用简单的代数将方程式右侧的未知元件移到方程式的左侧, 以解决元件 :

$\begin{aligned} 2 \int \cos (\ln x) d x & = x \cos (\ln x) + x \sin (\ln x) \\ \int \cos (\ln x) d x & = \frac{1}{2} x [\cos (\ln x) + \sin (\ln x)] + C \dots Use the same dummy constant of integration . \end{aligned}$

::2cos( lnxx) dx=xcos( lnx) +xsin( lnx) cos( lnx) dx=12x[ cos( lnx) +sin( lnx) +C... 使用相同的模拟整合常数。

A two-pass approach has been used to obtain the solution by getting back to the original integral.
::现已采用双向办法,回到原来的整体体,从而获得解决办法。

Examples
::实例

Example 1
::例1

Earlier, you were asked if you can solve for the integral $\begin{aligned} \int \cos 5 x \sin 7 x d x \end{aligned}$ .
::早些时候,有人问您能否解答 Cos5xsin7xdx。

Using $\begin{aligned} u = \cos 5 x, d v = \sin 7 x d x : \end{aligned}$
::使用 u=cos=5x,dv=sin=sin=7xdx:

1 ^st pass: $\begin{aligned} \int \cos 5 x \sin 7 x d x = - \frac{1}{7} \cos 5 x \cos 7 x - \frac{1}{35} \int \sin 5 x \cos 7 x d x \end{aligned}$
::第一关: cos5xsin7xdx17cos5xxcos7x-135sin5xxcos7xdx

Using $\begin{aligned} u = \sin 5 x, d v = \cos 7 x d x : \end{aligned}$
::使用 us=sin @% 5x, dv=cos @7xdx:

2 ^nd pass: $\begin{aligned} \int \cos 5 x \sin 7 x d x = - \frac{1}{7} \cos 5 x \cos 7 x - \frac{1}{35} [\frac{1}{7} \sin 5 x \sin 7 x + \frac{1}{35} \int \cos 5 x \sin 7 x d x] \end{aligned}$
::第二关: cos5xsin7xdx17cos5xxcos7x-135 [17sin5xsin7xxx7xx+135cos5xsin7xxxxx]

Note the integral $\begin{aligned} \int \cos 5 x \sin 7 x d x \end{aligned}$ on the right hand side of the equation?
::注意方程式右侧的积分 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

Solving for the integral:
::为集成体解决问题 :

$\begin{aligned} \int \cos 5 x \sin 7 x d x = \frac{35^{2}}{35^{2} + 1} [- \frac{1}{7} \cos 5 x \cos 7 x - \frac{1}{7 \cdot 35} \sin 5 x \sin 7 x] + C \end{aligned}$

::7xdxx=352352+1[ -17cos=5xcos=7xxxxxxxxxx]+C] [ -17cos=5xxxxxxxxx+C] [ -17cos=5xxxxxxxxxxxxxxxxx+C]

This is pretty messy. Later you will see a different way to handle products of sines and cosines with different angles.
::这是相当混乱的。以后你会看到一种不同的处理方式处理用不同角度的绳子和连结的产品。

Example 2
::例2

Evaluate $\begin{aligned} \int \sec^{3} x d x \end{aligned}$ .
::评估 sec3xdx。

Rewrite the integrand: $\begin{aligned} \int \sec^{3} x d x = \int \sec x \sec^{2} x d x \end{aligned}$
::重写整数( I) : @ sec3 @ xdx @ sec=xsec2 @ xdx

$\begin{aligned} Choose : u = \sec x d v = \sec^{2} x d x \\ Form : d u = \sec x \tan x d x v = \tan x \\ Evaluate : \int \sec^{3} x d x = \sec x \tan x - \int \sec x \tan^{2} x d x \end{aligned}$

::选择: u=secxdv=secxdxForm: du=secxtan}xdxv=tan}xvaluate:\\sec3}xdx=secxtan}xx@secx}xdx=secxtan}xtan2}xdxxxxx

We know that $\begin{aligned} \tan^{2} x \end{aligned}$ is related to $\begin{aligned} \sec x \end{aligned}$ : $\begin{aligned} \tan^{2} x = \sec^{2} x - 1 \end{aligned}$ , so that we have:
::我们知道, tan2x 与 secx 有关 : tan2 x=sec2 x-1, 所以我们有:

$\begin{aligned} \int \sec^{3} x d x & = \sec x \tan x - \int \sec x (\sec^{2} x - 1) d x \\ = \sec x \tan x - \int \sec^{3} x d x + \int \sec x d x \end{aligned}$

::sec3xdx=secxtanxsecxxxx(sec2_x_1)dx=secxtanx}sec3xdxsecxdxxx

Notice that the unknown integral now appears on both sides of the equation. We can use simple algebra to move the unknown integral on the right to the left side of the equation to solve for the integral:
::注意未知的元件现在出现在方程式的两侧。我们可以使用简单的代数将方程式右侧的未知元件移到方程式的左侧, 以解决元件 :

$\begin{aligned} 2 \int \sec^{3} x d x & = \sec x \tan x + \int \sec x d x \\ \int \sec^{3} x d x & = \frac{1}{2} [\sec x \tan x + \int \sec x d x] + C \\ = \frac{1}{2} [\sec x \tan x + \ln | \sec x + \tan x |] + C \end{aligned}$

::+C=12 [secáxtan]+C=12[secxtan]+C=12[secxtan]xxxxxl+lnçsecxx+tanx]+C]+C=12[secxtan]

Example 3
::例3

Evaluate $\begin{aligned} \int 3 (e^{- 2 x} + 1) \cos 4 x d x \end{aligned}$ .
::评估 3 (e-2x+1) cos 4xdx。

The problem integral can be transformed as follows:
::整体问题可以改变如下:

$\begin{aligned} \int 3 (e^{- 2 x} + 1) \cos 4 x d x & = 3 \int e^{- 2 x} \cos 4 x d x + 3 \int \cos 4 x d x \\ = 3 \int e^{- 2 x} \cos 4 x d x + \frac{3}{4} \sin 4 x + C \end{aligned}$

::3 (e- 2x+1) 3 (e- 2x+1) cos @4xxx=3 e- 2xcos=4xxx+3 cos=4xxxx=3 e- 2xcos=4xdx+34sin=4xx+C

The main problem here is to evaluate $\begin{aligned} \int e^{- 2 x} \cos 4 x d x \end{aligned}$ , which can be done using integration by parts.
::这里的主要问题是评估e-2xcos4xdx,这可以通过按部分的整合来完成。

$\begin{aligned} Use : \int u d v = u v - \int v d u . \dots 1^{s t} Pass \\ Choose : u = \cos 4 x d v = e^{- 2 x} d x \\ Form : d u = \frac{\sin 4 x}{4} d x v = - \frac{e^{- 2 x}}{2} \\ Evaluate : \int e^{- 2 x} \cos 4 x d x = - \frac{e^{- 2 x}}{2} \cos 4 x + \int - \frac{e^{- 2 x}}{2} \frac{\sin 4 x}{4} d x \\ = - \frac{e^{- 2 x}}{2} \cos 4 x - \frac{1}{8} \int e^{- 2 x} \sin 4 x d x \end{aligned}$

::使用 : @ udv=vuvvdu... 1passChoose: u=cos=4xdv=e-2xxForm: du=sin=4x4x4dxv=e_2x2x2Evaluate: @ e=2xcos=4xxxxxxx*2x2cos=4x2x2sin=4x4dxx @ e=2x2x2cos=4x4xdxxxxxxxxx

Notice that the second integral looks similar to our original integral in form, except that it has a $\begin{aligned} \sin 4 x \end{aligned}$ instead of $\begin{aligned} \cos 4 x \end{aligned}$ . To evaluate it, we again apply integration by parts to the second term:
::请注意,第二个整体体看起来与我们原来的整体体形式相似, 但它有一个sin4x而不是cos4x。要评估它, 我们再次对第二个术语部分地应用整合 :

$\begin{aligned} Use : \int u d v = u v - \int v d u . \dots 2^{n d} Pass \\ Choose : u = \sin 4 x d v = e^{- 2 x} d x \\ Form : d u = - \frac{\cos 4 x}{4} d x v = - \frac{e^{- 2 x}}{2} \\ Evaluate : \int e^{- 2 x} \cos 4 x d x = - \frac{e^{- 2 x}}{2} \cos 4 x + [- \frac{e^{- 2 x}}{2} \sin 4 x - \int (- \frac{e^{- 2 x}}{2}) (- \frac{\cos 4 x}{4}) d x] \\ = - \frac{e^{- 2 x}}{2} \cos 4 x - \frac{e^{- 2 x}}{2} \sin 4 x - \frac{1}{8} \int e^{- 2 x} \cos 4 x d x \end{aligned}$

::使用 : @ udv=vvvvdu.... 2nd PassChoose: u=sin @ 4xdv=e- 2xxForm: du @ cos=4x4dxv=e_ 2x2x2Evaluate: @ e=2xcos=4xxxx4x+[- e-2x2xsin4xx4xx}}( e2- 2x2x2)(- cos=4x4x4dx)] @ e- 2x2x2xxx_2xx_4x_18_e- 2xxx=4xxxxxxxxxxxxxxxxxxxx

The unknown integral now appears on both sides of the equation, and we can use simple algebra to move the unknown integral on the right to the left side of the equation to solve for the integral:
::未知的有机体现在出现在方程的两侧, 我们可以使用简单的代数将方程右侧的未知有机体移到左边,

$\begin{aligned} \frac{9}{8} \int e^{- 2 x} \cos 4 x d x & = - \frac{e^{- 2 x}}{2} \cos 4 x - \frac{e^{- 2 x}}{2} \sin 4 x \\ \int e^{- 2 x} \cos 4 x d x & = - \frac{4}{9} (\cos 4 x + \sin 4 x) e^{- 2 x} \end{aligned}$

::- 2xcos=4xdxxe - 2x2cos=4xe -2x2sin=4xxxx_2xxxcos=4xxxxx49(cos=4x+sin=4x)e-2x

Now, we can take this result and solve the original problem:
::现在,我们可以采取这个结果并解决最初的问题:

$\begin{aligned} \int 3 (e^{- 2 x} + 1) \cos 4 x d x & = 3 \int e^{- 2 x} \cos 4 x d x + 3 \int \cos 4 x d x \\ = 3 [- \frac{4}{9} (\cos 4 x + \sin 4 x) e^{- 2 x}] + \frac{3}{4} \sin 4 x + C \\ \int 3 (e^{- 2 x} + 1) \cos 4 x d x & = - \frac{4}{3} (\cos 4 x + \sin 4 x) e^{- 2 x} + \frac{3}{4} \sin 4 x + C \end{aligned}$

::3(e- 2x+1) 4xxx=32xcos4xxx+34xxxx=3[49(cos4x+sin4xx)e-2x] +34sin4x+C}3(e- 2x+1)cos4xdxxx=43(cos4x+sin4x)e-2x+34sin4xx+C

Review
::回顾

Evaluate the following integrals using integration by parts to solve for the unknown integral.
::利用各部分的整合来评估以下组合,以解决未知组合。

$\begin{aligned} \int e^{x} \sin 2 x d x \end{aligned}$
::============================================================================================================================================== ============================================================================================================================================================================================================

$\begin{aligned} \int e^{- 3 x} \sin x d x \end{aligned}$
::e- 3xsin xdx

$\begin{aligned} \int e^{a x} \cos b x d x \end{aligned}$
::

$\begin{aligned} \int_{0}^{\frac{π}{6}} e^{- x} \cos 3 x d x \end{aligned}$
::06e -xcos3xxxx

$\begin{aligned} \int_{0}^{\frac{π}{2}} e^{2 x} \sin x d x \end{aligned}$
::02 e2xsin xdx

$\begin{aligned} \int \sin 2 x \cos 3 x d x \end{aligned}$
::2xcos

$\begin{aligned} \int \sin 4 x \sin 3 x d x \end{aligned}$
::3xdx

$\begin{aligned} \int \cos 3 x \cos 4 x d x \end{aligned}$
::============================================================================================================================================= ==============================================================================================================================

$\begin{aligned} \int_{0}^{\frac{π}{2}} 4 \sin x \cos (\frac{x}{3}) d x \end{aligned}$
::024sinxcos( x3) dx

$\begin{aligned} \int_{0}^{\frac{π}{2}} 4 \sin (\frac{x}{3}) \cos (x) d x \end{aligned}$
::024sin(x3)cos(x)dx

$\begin{aligned} \int \sin (\ln x) d x \end{aligned}$
:lnx) dx

$\begin{aligned} \int \cos (\ln 0.5 x) d x \end{aligned}$
::============================================================================================================================================= ====================================================================== ========================================================================================

$\begin{aligned} \int \cos (2 \ln x) d x \end{aligned}$
::*cos( 2lnx) dx

$\begin{aligned} \int \tan (\ln x) d x \end{aligned}$
:lnx) dx

$\begin{aligned} \int \csc^{5} x d x \end{aligned}$ , Hint: let $\begin{aligned} u = \csc^{3} x \end{aligned}$ in the first pass and use $\begin{aligned} \cot^{2} x = \csc^{2} x - 1 \end{aligned}$ .
::csc5xdx, 提示 : 让我们在第一关使用 csc3x 并使用 comt2x = csc2x- 1 。

$\begin{aligned} \int \sin x \sinh x d x \end{aligned}$
::xinh xdx

Review (Answers)
::回顾(答复)

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。

Section outline

Solving for the Unknown Integral ::未知集成器的解决

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::回顾

Review (Answers) ::回顾(答复)

Solving for the Unknown Integral
::未知集成器的解决

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Review
::回顾

Review (Answers)
::回顾(答复)