Course: 8.5 将松树和古柯树的力量融合在一起

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将松树和古柯树的力量结合起来

The voltage associated with an alternating electric current in your home might have the form $\begin{align*}V(t)=A \sin \left(\frac{2 \pi}{T} t \right)\end{align*}$ volts ( $\begin{align*}T\end{align*}$ is the period). Because the voltage varies with time, a standard way to characterize the voltage in a single number is to compute the root-mean-square (rms) value. This means to square $\begin{align*}V(t)\end{align*}$ , take the mean or average over one period, then take the square root. Can you determine what the rms voltage would be if the voltage in your home could get as high as 155 volts?
::与您家中交替电流相关的电压可能具有表V(t)=Asin(2QTt)伏特(T是周期)。因为电压随时间变化而变化,用一个数字来描述电压的标准方式是计算根平方值。这意味着正方V(t),一个时期的平均值或平均值,然后取平方根。如果您家的电压可能高达155伏特,您能否确定Rms电压会是什么?

Integrating Powers of Sines and Cosines
::将松树和古柯树的力量结合起来

In this section we study methods of integrating functions of the form:
::在本节中,我们研究整合表格功能的方法:

$\begin{align*}\int \sin^m x \cos^n x dx\end{align*}$
::辛克辛克斯克斯xxxxx

where $\begin{align*}m\end{align*}$ , and $\begin{align*}n\end{align*}$ are nonnegative integers.
::其中 m 和 n 是非负整数。

The following three general categories or problem forms are of interest:
::以下三种一般类别或问题形式值得注意:

$\begin{align*}n=0\end{align*}$ , so that problems look like $\begin{align*}\int \sin^m x dx\end{align*}$ ;
::n=0,所以问题看起来像sinmxdx;

$\begin{align*}m=0\end{align*}$ , so that problems look like $\begin{align*}\int \cos^n x dx\end{align*}$ ;
::m=0,所以问题看起来像cosnxx;

$\begin{align*}m \ge 1\end{align*}$ and $\begin{align*}n \ge 1\end{align*}$ , so that problems look like $\begin{align*}\int \sin^m x \cos^n x dx\end{align*}$ .
::m1 和 n1, 所以问题看起来像 sinmxcosnxdx。

Let’s look at the three forms in order.
::让我们看看这三种形式。

Problem Form I: $\begin{align*}\int \sin^m x dx\end{align*}$
::问题表一:=============================================================================================================

This integral form can be evaluated using a general reduction formula (derived by applying integration by parts). A reduction formula is a formula solution that solves an integral problem by reducing it to a problem of solving an easier integral problem, which in turn can be solved by reducing the new integral to an easier problem, and so on. The reduction formula for the integral is:
::这种整体形式可以用一般的削减公式(通过按部分进行整合来得出)来评估。一种削减公式是一种解决一个整体问题的公式解决办法,通过将它简化为解决一个较容易的整体问题的办法来解决一个整体问题,而解决这个整体问题又可以通过减少一个较容易的问题的新整体来加以解决,等等。

Reduction Formula:
::减少公式:

$\begin{align*}\int \sin^m x dx=- \frac{1}{m} \sin^{m-1}x \cos x+ \frac{m-1}{m} \int \sin^{m-2}x dx\end{align*}$
::1msinm- 1xcosx+m- 1m%sinm- 2xxx

You can check that this formula works by trying it out for the case $\begin{align*}m=1\end{align*}$ , which leads to $\begin{align*}\int \sin^1 x dx=- \frac{1}{1} \sin^{1-1} x \cos x+ \frac{1-1}{1} \int \sin^{m-2}x dx=- \cos x\end{align*}$ .
::您可以通过测试大小写 m=1 来检查该公式是否有效, 因为它导致 {sin1xdx\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ </span> </p> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-NjY2YzZhNzc0ODY1MmMxMGI3ZDA1MmViZmFjM2ExZjA.-rnl"> This is a familiar result. <br/> <span style="color: green; "> ::这是一个熟悉的结果。 </span> </p> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-MDVjMzE2NTRmNzk2MzZiYzczNzQ3ZWNmNzFiNTIzYjU.-bgq"> As an alternative to using the reduction formula, the table below provides a guide to evaluating problems with this form based on whether the exponent is odd or even. <br/> <span style="color: green; "> ::作为使用减排公式的替代办法,下表提供了一种指南,用以根据指数是奇数还是偶数来评价这一表格的问题。 </span> </p> <table border="1" id="x-ck12-ZGRkMzdhYjNiMjg0YzYyZjk5MDU3NzkyOWYzMTFjMDc.-dx4"> <thead> <tr> <td> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-dyy"> <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-math" data-mathmethod="inline" data-tex="%5Cint%20%5Csin%5Em%20x%20dx"> <span class="MathJax_Preview" style="color: inherit;"> <span class="MJXp-math" id="MJXp-Span-244"> <span class="MJXp-mtable" id="MJXp-Span-245"> <span> <span class="MJXp-mtr" id="MJXp-Span-246" style="vertical-align: baseline;"> <span class="MJXp-mtd" id="MJXp-Span-247" style="text-align: right;"> <span class="MJXp-mo" id="MJXp-Span-248" style="margin-left: 0em; margin-right: 0.111em;"> <span class="MJXp-largeop MJXp-int"> ∫ </span> </span> <span class="MJXp-msubsup" id="MJXp-Span-249"> <span class="MJXp-mi" id="MJXp-Span-250" style="margin-right: 0.05em;"> sin </span> <span class="MJXp-mi MJXp-italic MJXp-script" id="MJXp-Span-251" style="vertical-align: 0.5em;"> m </span> </span> <span class="MJXp-mo" id="MJXp-Span-252" style="margin-left: 0em; margin-right: 0em;"> </span> <span class="MJXp-mi MJXp-italic" id="MJXp-Span-253"> x </span> <span class="MJXp-mi MJXp-italic" id="MJXp-Span-254"> d </span> <span class="MJXp-mi MJXp-italic" id="MJXp-Span-255"> x </span> </span> </span> </span> </span> </span> </span> <span class="MathJax_SVG MathJax_SVG_Processed" id="MathJax-Element-18-Frame" style="font-size: 100%; display: inline-block;" tabindex="-1"> <svg focusable="false" height="5.667ex" role="img" style="vertical-align: -2.241ex;" viewbox="0 -1474.9 5217.8 2440" width="12.119ex" xmlns:xlink="http://www.w3.org/1999/xlink"> <g fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"> <g transform="translate(167,0)"> <g transform="translate(-14,0)"> <use x="0" xlink:href="#MJSZ2-222B" y="0"> </use> <g transform="translate(1111,0)"> <use xlink:href="#MJMAIN-73"> </use> <use x="394" xlink:href="#MJMAIN-69" y="0"> </use> <use x="673" xlink:href="#MJMAIN-6E" y="0"> </use> <use transform="scale(0.707)" x="1738" xlink:href="#MJMATHI-6D" y="583"> </use> </g> <use x="3228" xlink:href="#MJMATHI-78" y="0"> </use> <use x="3801" xlink:href="#MJMATHI-64" y="0"> </use> <use x="4324" xlink:href="#MJMATHI-78" y="0"> </use> </g> </g> </g> </svg> </span> <script id="MathJax-Element-18" type="math/tex"> \begin{align*}\int \sin^m x dx\end{align*}
::

Procedure
::程序程序程序程序程序程序程序

Key Identities
::关键特征

$\begin{align*}m > 1\end{align*}$ is odd $\begin{align*}(m=2p+1)\end{align*}$
::m> 1 是奇数 (m=2p+1)

(1) Convert integral:
:1) 转换元件 :

$\begin{align*}\int \sin^m x dx=\int \sin^{2p} x \sin x dx\end{align*}$
::2 pxsinxdx

(2) Use the identity to convert integral:
:2) 使用身份转换元件:

$\begin{align*}\int \sin^{2p} x \sin x dx=\int (1- \cos^2 x)^p \sin x dx\end{align*}$
::*%sin2pxsinxdx *(1-cos2x) psinxdx

(3) Use $\begin{align*}u\end{align*}$ -substitution to change variables: $\begin{align*}u=\cos x\end{align*}$ and $\begin{align*}du=- \sin x dx\end{align*}$ yields
:3) 使用u替代来改变变量:u=cosx和dusinxdx 产量

$\begin{align*}\int (1- \cos^2 x)^p \sin x dx=- \int (1-u^2)^p du\end{align*}$
:1 - cos2x) psinxdx (1 - u2) pdu

(4) Expand the polynomial in $\begin{align*}u\end{align*}$ .
:4) 扩大在u.的多元性。

(5) Integrate the resulting polynomial in terms of $\begin{align*}u\end{align*}$ .
:5) 将由此产生的多元性结合到u。

(6) Substitute back $\begin{align*}\cos x\end{align*}$ for $\begin{align*}u\end{align*}$ .
:6) u.

$\begin{align*}\sin^2 x=(1- \cos^2 x)\end{align*}$
::exin2x=( 1 - cos2x)

$\begin{align*}m\end{align*}$ is even $\begin{align*}(m=2p)\end{align*}$
::m 是平平( m=2p)

(1) Convert the integral:
:1) 转换元件:

$\begin{align*}\int \sin^m x dx=\int \sin^{2p}x dx\end{align*}$
::2 pxdx

(2) Use the identity to convert the integral:
:2) 使用身份转换积分:

$\begin{align*}\int \sin^{2p}x dx=\int \left(\frac{1}{2}(1- \cos 2x) \right)^p dx\end{align*}$
::2pxdx( 12(1) - cos2x) pdx

(3) Expand in powers of $\begin{align*}\cos 2x\end{align*}$ and apply the odd/even procedure of Problem Form II as many times as necessary to simplify the integrand .
:3) 扩大cos2x的权力,并尽可能多地采用问题表二的奇/偶程序,以简化整数。

$\begin{align*}\sin^2 x=\frac{1}{2} (1- \cos 2x)\end{align*}$
::exin2x=12( 1 - cos2x)

Let's apply the formula to evaluate $\begin{align*}\int \sin^3 x dx\end{align*}$ .
::让我们应用公式来评价sin3xdx。

$\begin{align*}\int \sin^3 x dx&=\int \sin^2 x \sin x dx && \ldots \text{Separate to even and odd powers because} \ m=3, \ \text{odd}. \\ &=\int (1- \cos^2 x) \sin x dx && \ldots \text{Used the identity}. \ \sin^2 x=(1- \cos^2 x) \\ &=- \int (1-u^2)du && \ldots \text{Used substitution}: \ u=\cos x \ \text{and} \ du=- \sin x dx \\ &=- \left(u- \frac{u^3}{3} \right)+C && \ldots \text{Integrated} \\ \int \sin^3 x dx&=- \cos x+ \frac{\cos^3 x}{3}+C && \ldots \text{Substituted} \ \cos x \ \text{for} \ u.\end{align*}$
::辛3xxxxxin2xxxxxxxxx... 与偶数和奇数功率分离, 因为 m=3, 奇数。\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\c\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Problem Form II: $\begin{align*}\int \cos^n x dx\end{align*}$
::问题表二:=======================================================================================================

This integral form can also be evaluated using a general reduction formula (derived by applying integration by parts).
::这一整体形式也可以使用一般削减公式(通过按部分进行整合来得出)来评估。

Reduction Formula: $\begin{align*}\int \cos^n x dx\end{align*}$
::减少公式:cosnxx

$\begin{align*}\int \cos^n x dx=\frac{1}{n} \cos^{n-1}x \sin x+ \frac{n-1}{n} \int \cos^{n-2}x dx\end{align*}$
::cosnxx=1nconsn- 1xsinx+n- 1ncosn-2xdx

You can check that this formula works by trying it out for the case $\begin{align*}n=1\end{align*}$ , which leads to $\begin{align*}\int \cos^1 x dx=\frac{1}{1} \cos^{1-1}x \sin x+ \frac{1-1}{1} \int \cos^{n-2} x dx=\sin x\end{align*}$ .
::您可以通过尝试 vase n=1 来检查此公式是否有效, 因为它导致 {cos1xdx=11cos1_ 1xsinx+1- 11{cosn_ 2xdx=sinx 。 </span> </p> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-NjY2YzZhNzc0ODY1MmMxMGI3ZDA1MmViZmFjM2ExZjA.-eps"> This is a familiar result. <br/> <span style="color: green; "> ::这是一个熟悉的结果。 </span> </p> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-NDdkNTY3NzYwMGQ2YjEyZDNmMDZiN2NiYmVkYjUxMGI.-dj9"> Again, as an alternative to using the reduction formula, the table below provides a guide to evaluating problems with this form based on whether the exponent is odd or even. <br/> <span style="color: green; "> ::同样,作为使用减排公式的替代办法,下表提供了一种指南,用以根据指数是奇数还是偶数来评价这一表格的问题。 </span> </p> <table border="1" class="x-ck12-nofloat" id="x-ck12-ZjQxZmQxOWZjYzdkOGQ3ODdlYWQ3MWJmMGYwZWEyODU.-dhh"> <thead> <tr> <td> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-mzv"> <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-math" data-mathmethod="inline" data-tex="%5Cint%20%5Ccos%5En%20x%20dx"> <span class="MathJax_Preview" style="color: inherit;"> <span class="MJXp-math" id="MJXp-Span-860"> <span class="MJXp-mtable" id="MJXp-Span-861"> <span> <span class="MJXp-mtr" id="MJXp-Span-862" style="vertical-align: baseline;"> <span class="MJXp-mtd" id="MJXp-Span-863" style="text-align: right;"> <span class="MJXp-mo" id="MJXp-Span-864" style="margin-left: 0em; margin-right: 0.111em;"> <span class="MJXp-largeop MJXp-int"> ∫ </span> </span> <span class="MJXp-msubsup" id="MJXp-Span-865"> <span class="MJXp-mi" id="MJXp-Span-866" style="margin-right: 0.05em;"> cos </span> <span class="MJXp-mi MJXp-italic MJXp-script" id="MJXp-Span-867" style="vertical-align: 0.5em;"> n </span> </span> <span class="MJXp-mo" id="MJXp-Span-868" style="margin-left: 0em; margin-right: 0em;"> </span> <span class="MJXp-mi MJXp-italic" id="MJXp-Span-869"> x </span> <span class="MJXp-mi MJXp-italic" id="MJXp-Span-870"> d </span> <span class="MJXp-mi MJXp-italic" id="MJXp-Span-871"> x </span> </span> </span> </span> </span> </span> </span> <span class="MathJax_SVG MathJax_SVG_Processing" id="MathJax-Element-45-Frame" style="font-size: 100%; display: inline-block;" tabindex="-1"> </span> <script id="MathJax-Element-45" type="math/tex"> \begin{align*}\int \cos^n x dx\end{align*}
::cosnxxx

Procedure
::程序程序程序程序程序程序程序

Key Identities
::关键特征

$\begin{align*}n\end{align*}$ is odd $\begin{align*}(n=2p+1)\end{align*}$
::n 是奇数(n=2p+1)

(1) Convert integral:
:1) 转换元件 :

$\begin{align*}\int \cos^n x dx=\int \cos^{2p} x \cos x dx\end{align*}$
::2pxcosxxxxx

(2) Use the identity to convert integral:
:2) 使用身份转换元件:

$\begin{align*}\int \cos^{2p} x \cos x dx=\int (1- \sin^2 x)^p \cos x dx\end{align*}$
::*cos2pxcosxxdx *(1- 辛2x) pcosxdx

(3) Use $\begin{align*}u\end{align*}$ -substitution to change variables: $\begin{align*}u=\sin x\end{align*}$ and $\begin{align*}du=\cos x dx\end{align*}$ yields
:3) 使用 u 替代来改变变量:u=sinx 和 du=cosxdx 产量

$\begin{align*}\int (1- \sin^2 x)^p \cos x dx=\int (1-u^2)^p du\end{align*}$
:1- 辛2x) pcosxdx (1- u2) pdu

(4) Expand the polynomial in $\begin{align*}u\end{align*}$ .
:4) 扩大在u.的多元性。

(5) Integrate the resulting polynomial in terms of $\begin{align*}u\end{align*}$ .
:5) 将由此产生的多元性结合到u。

(6) Substitute back $\begin{align*}\sin x\end{align*}$ for $\begin{align*}u\end{align*}$ .
:6) u. y. y. y. y. y. y. y. y. y. e. y. y. e. y. e. y. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e.

$\begin{align*}\cos^2 x=(1- \sin^2 x)\end{align*}$
::COs2x=( 1 - 辛2x)

$\begin{align*}n\end{align*}$ is even $\begin{align*}(n=2p)\end{align*}$
::n 是偶数 (n=2p)

(1) Convert the integral:
:1) 转换元件:

$\begin{align*}\int \cos^n x dx=\int \cos^{2p}x dx\end{align*}$
::cosnxxxcos2pxdx

(2) Use the identity to convert the integral:
:2) 使用身份转换积分:

$\begin{align*}\int \cos^{2p}x dx=\int \left(\frac{1}{2}(1+ \cos 2x) \right)^p dx\end{align*}$
::cos2pxdx( 12(1 +cos2x)) pdx

(3) Expand the integrand to get powers of $\begin{align*}\cos 2x\end{align*}$ and apply the odd/even procedure as many times as necessary to simplify the integrand.
:3) 扩大指数以获得COS2x的功率,并尽可能多地采用奇/偶程序来简化Cos2x的功率。

$\begin{align*}\cos^2 x=\frac{1}{2}(1+ \cos 2x)\end{align*}$
::cos2x=12(1+cos2x)

Let's apply the formula to evaluate $\begin{align*}\int \cos^4 x dx\end{align*}$ .
::让我们应用公式来评价 cos4xdx。

$\begin{align*}\int \cos^4 x dx &=\int(\cos^2 x)^2 dx && \ldots n=4, \ \text{even}. \\ &=\int \left(\frac{1}{2}(1+ \cos 2x) \right)^2 dx && \ldots \text{Used the identity}: \ \cos^2 x=\frac{1}{2}(1+ \cos 2x) \\ &=\left(\frac{1}{2} \right)^2 \int (1+2 \cos 2x+ \cos^2 2x)dx && \ldots \text{Expanded the square} \\ &=\left(\frac{1}{2} \right)^2 \int \left(1+2 \cos 2x+ \frac{1}{2}(1+ \cos 4x) \right)dx && \ldots \text{Used the identity} \ \cos^2 x=\frac{1}{2}(1+ \cos 2x) \\ &=\left(\frac{1}{4} \right) \int \left(\frac{3}{2}+2 \cos 2x+ \frac{1}{2} \cos 4x \right) dx \\ &=\left(\frac{1}{4} \right) \left(\frac{3}{2}x+ \sin 2x+ \frac{1}{8} \sin 4x \right)+C && \ldots \text{Integrated term-by-term} \\ \int \cos^4 x dx &=\frac{3}{8}x+ \frac{1}{4} \sin 2x+ \frac{1}{32} \sin 4x+C\end{align*}$
::{cos4xx}}{{(cos2x)2dx}}}}}}}{(even.}}}}}}}}}使用身份 : =12=12(1+cos2x)=12(1+cos2x)=12(12)2dx=12(1+2c2x)xxx...* {(1+2co2x)2x+12(1+2x)dx... 使用身份 CO2x=12(1+cos2x)=14}}(32+2cos2x)+12cos4x)x=(14) (32x+sin2x+sin2x+18sin4x)+C...(14)\(32xxxxxx=38x=38x+14sin2x+132sin4x+C

Problem Form III: $\begin{align*}\int \sin^m x \cos^n x dx\end{align*}$ , $\begin{align*}m \ge 1\end{align*}$ and $\begin{align*}n \ge 1\end{align*}$
::问题三:%sinmxcosnxxx, m1和 n1

The table below provides a guide to evaluating problems with this form.
::下表为评估这种形式问题提供了指南。

$\begin{align}\int \sin^m x \cos^n x dx\end{align}$ ::辛克辛克斯克斯xxxxx	Procedure ::程序程序程序程序程序程序程序	Key Identities ::关键特征
$\begin{align}m\end{align}$ is odd $\begin{align}(m=2p+1)\end{align}$ ::m 是奇数 (m=2p+1)	(1) Convert integral: :1) 转换元件 : $\begin{align}\int \sin^m x \cos^n x dx=\int \sin^{2p} x \cos x \sin x dx\end{align}$ ::2pxcosxxxxxxxxxxxxxx (2) Use the identity to convert integral: :2) 使用身份转换元件: $\begin{align}\int \sin^{2p} x \cos^n x \sin x dx=\int(1- \cos^2 x)^p \cos^n x \sin x dx\end{align}$ ::2pxcosnxinxxxxxxx ( 1- cos2x) pccosnxinxxxxxxx (3) Use $\begin{align}u\end{align}$ -substitution to change variables: $\begin{align}u=\cos x\end{align}$ and $\begin{align}du=- \sin x dx\end{align}$ yields :3) 使用u替代来改变变量:u=cosx和dusinxdx 产量 $\begin{align}\int(1- \cos^2 x)^p \cos^n x \sin x dx = - \int (1-u^2)^p u^n du\end{align}$ :1- cos2x) pcosnxinxxdx (1- u2) pundu (4) Expand the product of $\begin{align}u^n\end{align}$ and the polynomial in $\begin{align}u\end{align}$ . :4) 在u中扩大非和多民族产物。 (5) Integrate the resulting polynomial in terms of $\begin{align}u\end{align}$ . :5) 将由此产生的多元性结合到u。 (6) Substitute back $\begin{align}\cos x\end{align}$ for $\begin{align}u\end{align}$ . :6) u.	$\begin{align}\sin^2 x=(1- \cos^2 x)\end{align}$ ::exin2x=( 1 - cos2x)
$\begin{align}n\end{align}$ is odd $\begin{align}(n=2p+1)\end{align}$ ::n 是奇数 (n=2p+1)	(1) Convert integral: :1) 转换元件 : $\begin{align}\int \sin^m x \cos^n x dx=\int \sin^m x \cos^{2p} x \cos x dx\end{align}$ ::2pxcosxxxxxx (2) Use the identity to convert integral: :2) 使用身份转换元件: $\begin{align}\int \sin^m x \cos^{2p} x \cos x dx=\int \sin^m x \cdot (1- \sin^2 x)^p \cos x dx\end{align}$ :: (3) Use $\begin{align}u\end{align}$ -substitution to change variables: $\begin{align}u=\sin x\end{align}$ and $\begin{align}du=\cos x dx\end{align}$ yields :3) 使用 u 替代来改变变量:u=sinx 和 du=cosxdx 产量 $\begin{align}\int \sin^m x \cdot (1- \sin^2 x)^p \cos x dx = \int u^m (1-u^2)^p du\end{align}$ :1- 辛2x) pccosxdx (1- u2) pdu (4) Expand the product of $\begin{align}u^m\end{align}$ and the polynomial in $\begin{align}u\end{align}$ . :4) 在u中扩大微米和多面体的产物。 (5) Integrate the resulting polynomial in terms of $\begin{align}u\end{align}$ . :5) 将由此产生的多元性结合到u。 (6) Substitute back $\begin{align}\sin x\end{align}$ for $\begin{align}u\end{align}$ . :6) u. y. y. y. y. y. y. y. y. y. e. y. y. e. y. e. y. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e.	$\begin{align}\cos^2 x=(1- \sin^2 x)\end{align}$ ::COs2x=( 1 - 辛2x)
$\begin{align}m\end{align}$ and $\begin{align}n\end{align}$ are even $\begin{align}(m=2p, n=2r)\end{align}$ ::m 和 n 是偶数 (m=2p,n=2r)	(1) Convert the integral: :1) 转换元件: $\begin{align}\int \sin^m x \cos^n x dx=\int \sin^{2p} x \cos^{2r} x dx\end{align}$ ::2pxcos2rxdx (2) Use the half-angle identities to reduce the exponents by one-half and convert the integral: :2) 使用半角身份将指数减少一半,并转换积分: $\begin{align}\int \sin^{2p} x \cos^{2r} x dx= \int \left(\frac{1}{2}(1- \cos 2x) \right)^p \left(\frac{1}{2}(1+ \cos 2x) \right)^r\end{align}$ (3) Expand the integrand to get powers of $\begin{align}\cos 2x\end{align}$ and apply the odd/even procedure as many times as necessary to simplify the integrand. ::=============================================================================	$\begin{align}\sin^2 x &=\frac{1}{2}(1- \cos 2x) \\ \cos^2 x &=\frac{1}{2}(1+ \cos 2x)\end{align}$ ::exi2x=12(1 - cos2x)cos2x=12(1+cos2x)

Let's apply the formula to evaluate $\begin{align*}\int \sin^3 x \cos^4 x dx\end{align*}$
::让我们应用公式来评价 sin3xcos4xdx

$\begin{align*}\int \sin^3 x \cos^4 x dx &=\int \sin^2 x \cos^4 x \sin x dx && \ldots m=3, \ \text{odd}. \\ &=\int(1- \cos^2 x) \cos^4 x \sin x dx && \ldots \text{Used the identity} \ \sin^2 x=(1- \cos^2 x) \\ &=- \int (1-u^2)u^4 du && \ldots \text{Used substitution}: \ u=\cos x \ \text{and} \ du=- \sin x dx \\ &=- \int (u^4-u^6)du \\ &=- \frac{u^5}{5}+ \frac{u^7}{7}+C && \ldots \text{Integrated} \\ \int \sin^3 x \cos^4 x dx &=- \frac{\cos^5 x}{5}+ \frac{\cos^7 x}{7}+C && \ldots \text{Substituted} \ \cos x \ \text{for} \ u.\end{align*}$
::辛3xcos4xxxxxxxxxxxxxxxx...m=3, 奇数。\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ xxxxxxxxxxxxxxxxxxxxxxxxx\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Examples
::实例

Example 1
::例1

Earlier, you were asked to determine what the root-mean-square(rms) voltage would be if the voltage in your home could get as high as 155 volts.
::早些时候,有人要求你确定,如果家里的电压高达155伏,那么根平方(rms)的电压会是多少。

The procedure for computing the rms voltage is to square $\begin{align*}V(t)=A \sin \left(\frac{2 \pi}{T}t \right)\end{align*}$ , take the mean or average over one period, then take the square root. This means that the root-mean-square voltage can be computed as:
::计算 rms 电压的程序是 qua V (t) = Asin (2QTt), 以一个时期的平均值或平均值取一个时期, 然后取平方根。这意味着根平方电压可以以下列方式计算:

$\begin{align*}\sqrt{\frac{1}{T} \int\limits_{0}^{T} V^2(t)dt}= \sqrt{\frac{1}{T} \int\limits_{0}^{T} A^2 \sin^2 \left(\frac{2 \pi}{T}t \right)dt}=\sqrt{\frac{A^2}{2 T} \int\limits_{0}^{T} \left[1- \cos \left(\frac{4 \pi}{T}t \right) \right]}dt=\sqrt{\frac{A^2}{2 T}T}=\frac{\sqrt{2}}{2}A\end{align*}$ , where the double angle relationship $\begin{align*}\sin^2 x=\frac{1}{2}(1- \cos 2x)\end{align*}$ is used.
::1TT0V2(t)dt1T0A2sin2(2QT)dA22T0[1-cos(4QTt)]dtA22T22A, 此处使用双角关系 sin2x=12(1-cos2x) 。

Therefore, if $\begin{align*}A=155\end{align*}$ volts, then the rms voltage is 110 volts. This is a familiar number.
::因此,如果A=155伏,Rms电压为110伏。这是一个常见的数字。

Example 2
::例2

Evaluate $\begin{align*}\int \sin^4 x \cos^4 x dx\end{align*}$ .
::评估============================================================================================================================================= ==============================================================================================================================================================================================================================================================================================================================================================================

Here, $\begin{align*}m=n=4\end{align*}$ We follow the third procedure in the table above:
::在这里, m=n=4 我们遵循上表的第三种程序:

$\begin{align*}\int \sin^4 x \cos^4 x dx &=\int (\sin^2 x)^2 (\cos^2 x)^2 dx && \ldots m, n \ \text{are even}. \\ & && \ldots \text{Used the identities}: \\ & =\int \left(\frac{1}{2}(1- \cos 2x) \right)^2 \left(\frac{1}{2}(1+ \cos 2x) \right)^2 dx &&\sin^2 x=\frac{1}{2}(1- \cos 2x) \\ & && \cos^2 x=\frac{1}{2}(1+ \cos 2x) \\ &=\left(\frac{1}{2} \right)^4 \int (1- \cos^2 2x)^2 dx && \ldots \text{Simplified the integrand} \\ &=\left(\frac{1}{16} \right) \int (\sin^2 2x)^2 dx && \ldots \text{Used the identity} \ 1- \cos^2 x=\sin^2 x \\ &=\left(\frac{1}{16} \right) \int \left(\frac{1}{2}(1- \cos 4x) \right)^2 dx && \ldots \text{Used the identities}: \ \sin^2 2x=\frac{1}{2}(1- \cos 4x) \\ &=\left(\frac{1}{64} \right) \int (1-2 \cos 4x+ \cos^2 4x)dx && \ldots \text{Expanded the integrand} \\ &=\left(\frac{1}{64} \right) \int \left(1- 2 \cos 4x+ \frac{1}{2}(1+ \cos 8x) \right)dx \\ &=\left(\frac{1}{64} \right) \int \left(\frac{3}{2}-2 \cos 4x+ \frac{\cos 8x}{2} \right)dx \\ &=\left(\frac{1}{64} \right) \left(\frac{3x}{2}- \frac{\sin 4x}{2}+ \frac{\sin 8x}{16} \right)+C && \ldots \text{Integrated terms} \\ \int \sin^4 x \cos^4 x dx &=\frac{3x}{128}- \frac{\sin 4x}{128}+ \frac{\sin 8x}{1024}+C\end{align*}$
::=sin4xCos4xxxxxx)2,228x2dxx... 简化了 Integrand= (116) (1) (2) (c2x2x) 2,n是ee. 使用身份 :\ (12(1) (1) (1) (1) (1) (1) (2) co2x, 2xxxxx 。身份 : (12) (12) (1) (1) (1) (1) 4) 4(1) 4(1) 4x... 简化了 Ingrand= (116)(2) (1) (1) (1) (1) (2) (1) (2 4) (1) (1) (2 (4) (4) (4)(2 (4)(2) (4) (4)(2 (co 4) (4) (4) (4) ([2 (4) (4, (4) (4(4) (4) (4) (4(4) (4) (4(4) (4) (4) (4(4) (4(4) (4) (4) (4(4) (4(4(4) (4) (4) (4(4) (4) (4) (4) (4) (4) (4) (4) (4) (4(4) (4) (4) (4) (4) (4) (4) (4(4) (4) (4) (4) (4) (4) (4(4) (4) (4) (4) (4) (4(4(4(4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4(4(4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4(4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4) (4) ((4) (4) (

Review
::回顾

Evaluate the integrals.
::评估整体体。

$\begin{align*}\int \sin^2 (x)dx\end{align*}$
::sin2(x)dx

$\begin{align*}\int \sin^2 5 \phi d \phi\end{align*}$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}新年25日 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}新年25日 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}是的 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}是的 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}是的 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}是的

$\begin{align*}\int \cos^2 (x)dx\end{align*}$
::*cos2(x)dx

$\begin{align*}\int \cos^3 (x)dx\end{align*}$
::*cos3 (x) dx

$\begin{align*}\int \sin^4 (x)dx\end{align*}$
::辛 4 (x) dx

$\begin{align*}\int \sin^5 (x)dx\end{align*}$
::辛5 (x) dx

$\begin{align*}\int \cos^4 x \sin x dx\end{align*}$
::cos4xsinxdxxx cos4xsinxxxxxx cos4xsinxxxxxxx

$\begin{align*}\int \sin^2 2z \cos^3 2 z dz\end{align*}$
::zcoscos 32zdzzz

$\begin{align*}\int z^2 \sin^2 2z^3 \cos^3 2z^3 dz\end{align*}$
::z2sin22z3cos32z3dz z2sin22z3cos32z3dz

$\begin{align*}\int \sin^5 (x) \cos^4 (x) dx\end{align*}$
::sin5(x)cos4(x)dx

$\begin{align*}\int \sin^3 (x) \cos^2 (x) dx\end{align*}$
::==================xxxxxxxxxxxxxxxx==========================xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\begin{align*}\int \sin (x) \cos^2 (x) dx\end{align*}$
::sin( x) cos2( x) dx

$\begin{align*}\int \sin^2 (x) \cos (x) dx\end{align*}$
::sin2(x)cos(x)dx

$\begin{align*}\int \sin^2 (x) \cos^2 (x) dx\end{align*}$
::*sin2(x)cos2(x)dx ======================================xxxxdxxxxxxxxxxxxxxxxx===================sin2(xxxxxdxxxxxxxxxxxxxxxxxxx=====================xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\begin{align*}\int \sin^4 (x) \cos^3 (x) dx\end{align*}$
::辛4 (x) cos3 (x) dx

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。

$\begin{align}\int \sin^m x \cos^n x dx\end{align}$ ::辛克辛克斯克斯xxxxx	Procedure ::程序程序程序程序程序程序程序	Key Identities ::关键特征
$\begin{align}m\end{align}$ is odd $\begin{align}(m=2p+1)\end{align}$ ::m 是奇数 (m=2p+1)	(1) Convert integral: :1) 转换元件 : $\begin{align}\int \sin^m x \cos^n x dx=\int \sin^{2p} x \cos x \sin x dx\end{align}$ ::2pxcosxxxxxxxxxxxxxx (2) Use the identity to convert integral: :2) 使用身份转换元件: $\begin{align}\int \sin^{2p} x \cos^n x \sin x dx=\int(1- \cos^2 x)^p \cos^n x \sin x dx\end{align}$ ::2pxcosnxinxxxxxxx ( 1- cos2x) pccosnxinxxxxxxx (3) Use $\begin{align}u\end{align}$ -substitution to change variables: $\begin{align}u=\cos x\end{align}$ and $\begin{align}du=- \sin x dx\end{align}$ yields :3) 使用u替代来改变变量:u=cosx和dusinxdx 产量 $\begin{align}\int(1- \cos^2 x)^p \cos^n x \sin x dx = - \int (1-u^2)^p u^n du\end{align}$ :1- cos2x) pcosnxinxxdx (1- u2) pundu (4) Expand the product of $\begin{align}u^n\end{align}$ and the polynomial in $\begin{align}u\end{align}$ . :4) 在u中扩大非和多民族产物。 (5) Integrate the resulting polynomial in terms of $\begin{align}u\end{align}$ . :5) 将由此产生的多元性结合到u。 (6) Substitute back $\begin{align}\cos x\end{align}$ for $\begin{align}u\end{align}$ . :6) u.	$\begin{align}\sin^2 x=(1- \cos^2 x)\end{align}$ ::exin2x=( 1 - cos2x)
$\begin{align}n\end{align}$ is odd $\begin{align}(n=2p+1)\end{align}$ ::n 是奇数 (n=2p+1)	(1) Convert integral: :1) 转换元件 : $\begin{align}\int \sin^m x \cos^n x dx=\int \sin^m x \cos^{2p} x \cos x dx\end{align}$ ::2pxcosxxxxxx (2) Use the identity to convert integral: :2) 使用身份转换元件: $\begin{align}\int \sin^m x \cos^{2p} x \cos x dx=\int \sin^m x \cdot (1- \sin^2 x)^p \cos x dx\end{align}$ :: (3) Use $\begin{align}u\end{align}$ -substitution to change variables: $\begin{align}u=\sin x\end{align}$ and $\begin{align}du=\cos x dx\end{align}$ yields :3) 使用 u 替代来改变变量:u=sinx 和 du=cosxdx 产量 $\begin{align}\int \sin^m x \cdot (1- \sin^2 x)^p \cos x dx = \int u^m (1-u^2)^p du\end{align}$ :1- 辛2x) pccosxdx (1- u2) pdu (4) Expand the product of $\begin{align}u^m\end{align}$ and the polynomial in $\begin{align}u\end{align}$ . :4) 在u中扩大微米和多面体的产物。 (5) Integrate the resulting polynomial in terms of $\begin{align}u\end{align}$ . :5) 将由此产生的多元性结合到u。 (6) Substitute back $\begin{align}\sin x\end{align}$ for $\begin{align}u\end{align}$ . :6) u. y. y. y. y. y. y. y. y. y. e. y. y. e. y. e. y. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e. e.	$\begin{align}\cos^2 x=(1- \sin^2 x)\end{align}$ ::COs2x=( 1 - 辛2x)
$\begin{align}m\end{align}$ and $\begin{align}n\end{align}$ are even $\begin{align}(m=2p, n=2r)\end{align}$ ::m 和 n 是偶数 (m=2p,n=2r)	(1) Convert the integral: :1) 转换元件: $\begin{align}\int \sin^m x \cos^n x dx=\int \sin^{2p} x \cos^{2r} x dx\end{align}$ ::2pxcos2rxdx (2) Use the half-angle identities to reduce the exponents by one-half and convert the integral: :2) 使用半角身份将指数减少一半,并转换积分: $\begin{align}\int \sin^{2p} x \cos^{2r} x dx= \int \left(\frac{1}{2}(1- \cos 2x) \right)^p \left(\frac{1}{2}(1+ \cos 2x) \right)^r\end{align}$ (3) Expand the integrand to get powers of $\begin{align}\cos 2x\end{align}$ and apply the odd/even procedure as many times as necessary to simplify the integrand. ::=============================================================================	$\begin{align}\sin^2 x &=\frac{1}{2}(1- \cos 2x) \\ \cos^2 x &=\frac{1}{2}(1+ \cos 2x)\end{align}$ ::exi2x=12(1 - cos2x)cos2x=12(1+cos2x)

Section outline

General

将松树和古柯树的力量结合起来

Integrating Powers of Sines and Cosines ::将松树和古柯树的力量结合起来

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Review ::回顾