Course: 8.6 将占地人和唐恩特人的权力结合起来

Section outline

Select section General

Collapse Expand
General

Collapse all Expand all
- Select activity 新闻通告
  
  新闻通告 Forum

将占地人和唐恩特人的权力结合起来

In many applications, particularly those involving arc length and surface area, one needs to evaluate integrals of the general form $\begin{align*}\int \tan^{m}x \sec^{n}xdx\end{align*}$ , where $\begin{align*}m\end{align*}$ and $\begin{align*}n\end{align*}$ are positive integers. In the previous concept on integrating , some useful integration rules were given based the identity $\begin{align*}\sin^{2}x+\cos^{2}x=1\end{align*}$ , the derivative relationships between the sine and cosine functions, and application of simple $\begin{align*}u\end{align*}$ -substitution or integration by parts techniques. In a similar way, useful guidelines for integrating powers of secants and tangents are derived by using their identity and derivative relationships. Before proceeding, see if you can write down the identity relationship between the tangent and secant functions (can you derive it using the sine/cosine relationship?). Do you know their derivative relationships? Can you evaluate $\begin{align*}\int\tan x \sec^{2}xdx\end{align*}$ ?
::在许多应用中,尤其是涉及弧长度和表面区域的应用中,人们需要评估普通表xsecnxdx(m和n是正整数)的组成部分。在先前的整合概念中,给出了一些有用的整合规则,其依据是身份 sin2x+cos2x=1,正弦函数和正弦函数之间的衍生关系,以及应用简单的替代技术或部件技术的集成关系。同样地,通过使用它们的身份和衍生关系,可以得出分离者和正切体权力整合的有用指南。在继续之前,请看您能否写下正弦函数和正弦函数之间的身份关系(您能否用正弦/正弦关系来得出?) 。您知道它们的衍生关系吗? 您能够评估 tanxse2xxxxxxxxxxx?

Integrating Powers of Secants and Tangents
::将占地人和唐恩特人的权力结合起来

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:

$\begin{align*}\int \tan^{m}x \sec^{n}xdx\end{align*}$
::{\fn方正黑体简体\fs18\b1\bord1\shad1\3cH2F2F2F} xsecn xdx

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 of problems are of interest:
::以下三大类问题值得注意:

$\begin{align*}n=0\end{align*}$ , so that problems look like $\begin{align*}\int \tan^{m}xdx\end{align*}$ ;
::n=0,所以问题看起来就像 nemxdx;

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

$\begin{align*}m\ge 1\end{align*}$ and $\begin{align*}n\ge 1\end{align*}$ , so that problems look like $\begin{align*}\int \tan^{m}x \sec^{n}xdx\end{align*}$ .
::m1 和 n1, 所以问题看起来就像 xsecn xdx。

Problem Form I: $\begin{align*}\int\tan^{m}xdx\end{align*}$

::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: $\begin{align*}\int\tan^{m}xdx,m\ge 2\end{align*}$
::减量公式:% tanmxdx, m2

$\begin{aligned} \int \tan^{m} x d x = \frac{\tan^{m - 1} x}{m - 1} - \int \tan^{m - 2} x d x \end{aligned}$

$\begin{align*}\int\tan^{m}xdx=\frac{\tan^{m-1}x}{m-1}-\int\tan^{m-2}xdx\end{align*}$
::

Note that the reduction formula applies for $\begin{align*}m\ge 2\end{align*}$ . But what is the result for $\begin{align*}m=1\end{align*}$ ?
::注意减量公式适用于 m2 。但是 m=1 结果是什么 ?

For the case of $\begin{align*}m=1\end{align*}$ , the integral is evaluated as follows:
::对于 m=1, 整体体评估如下:

\begin{aligned} \int \tan x d x & = \int \frac{\sin x}{\cos x} d x \\ = - \int \frac{1}{u} d u \\ = - \ln | u | + C \\ \int \tan x d x & = - \ln | \cos x | + C \end{aligned}

$\begin{align*}\int\tan x dx & = \int\frac{\sin x}{\cos x} dx\\ & = -\int\frac{1}{u}du\\ & = -\ln|u|+C\\ \int\tan x dx & = -\ln|\cos x|+C\end{align*}$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不...

For cases where $\begin{align*}m\ge 2\end{align*}$ , the table below provides an alternative guide to using the reduction formula for evaluating the form $\begin{align*}\int\tan^{m}xdx\end{align*}$ .
::对于m2的情况,下表提供了使用减排公式评价表tanmxdx的替代指南。

$\begin{align}\int\tan^{m}xdx,m\ge 2\end{align}$	Procedure ::程序程序程序程序程序程序程序	Key Identities ::关键特征
$\begin{align}m\end{align}$ is odd ::m是奇数,m是奇数 $\begin{align}(m=2p+1)\end{align}$ :m=2p+1)	(1) Convert integral: :1) 转换元件 : $\begin{align}\int\tan^{m}x dx=\int\tan^{2p}x\tan xdx\end{align}$ ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不,不,不 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不 (2) Use the identity to convert integral: :2) 使用身份转换元件: $\begin{align}\int\tan^{2p}x\tan x dx=\int(\sec^{2}x-1)^{p}\tan x dx\end{align}$ ::tan2pxtanxdx(sec2x- 1)ptanxdx (3) Use $\begin{align}u\end{align}$ -substitution to change variables: $\begin{align}u=\sec x\end{align}$ and $\begin{align}du=\sec x\tan x dx\end{align}$ yields :3) 使用 u 替代来改变变量: u=secx 和 du=secxtanxdx 产量 $\begin{align}\int(\sec^{2}x-1)^{p}\tan x dx=\int(u^{2}-1)^{p}\frac{du}{u}\end{align}$ :sec2x -1)pduu (4) Divide the polynomial function in $\begin{align}u\end{align}$ by $\begin{align}u\end{align}$ . :4) u除以 u 的多元函数除以 u。 (5) Integrate the resulting terms of $\begin{align}u\end{align}$ . :5) 将由此产生的u条款纳入其中。 (6) Substitute back $\begin{align}\sec x\end{align}$ for $\begin{align}u\end{align}$ . :6) 以 u 取代 secx。	$\begin{align}\tan^{2}x=\sec^{2}x-1\end{align}$ ::tan2\\\ x=sec2\\\\\\ x- 1 $\begin{align}u=\sec x\end{align}$ and $\begin{align}du=\sec x \tan x dx\end{align}$ , or $\begin{align}\frac{du}{u}=\tan x dx\end{align}$ ::u=secx 和 du=secxtanxdx, 或 duu=tanxdx
$\begin{align}m\end{align}$ is even ::以米为偶数 $\begin{align}(m=2p)\end{align}$ :m=2p)	(1) Convert the integral: :1) 转换元件: $\begin{align}\int\tan^{m}x dx=\int\tan^{2p}xdx\end{align}$ ::\ tanmxdxtan2pxdx (2) Use the identity to convert the integral: :2) 使用身份转换积分: $\begin{align}\int\tan^{2p}x dx=\int(\sec^{2}x-1)^{p} dx\end{align}$ ::\ tan2pxdx( sec2x- 1) pdx (3) Expand the integrand in powers (even) of $\begin{align}\sec x\end{align}$ . :3) 扩大cEZX的实权(甚至)范围。 (4) Integrate term by term the polynomial of even powers of $\begin{align}\sec x\end{align}$ using the reduction formula: :4) 使用削减公式的cEx偶数权力的多等式综合术语: $\begin{align}\int\sec^{n}xdx=\frac{\sec^{n-2}x \tan x}{n-1}+\frac{n-2}{n-1}\int \sec^{n-2}xdx\end{align}$ ::@ secnxdx=secn-2xtanxn-1+n-2n- 1secn-2xdx	$\begin{align}\tan^{2}x=(\sec^{2}x-1)\end{align}$ ::tan2_%x=( sec2_%x- 1)

\begin{aligned} \int \tan^{m} x d x, m \geq 2 \end{aligned}

$\begin{align*}\int\tan^{m}xdx,m\ge 2\end{align*}$

Procedure
::程序程序程序程序程序程序程序

Key Identities
::关键特征

$\begin{align*}m\end{align*}$ is odd
::m是奇数,m是奇数

$\begin{align*}(m=2p+1)\end{align*}$
:m=2p+1)

(1) Convert integral:
:1) 转换元件 :

$\begin{align*}\int\tan^{m}x dx=\int\tan^{2p}x\tan xdx\end{align*}$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不,不,不 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不

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

$\begin{align*}\int\tan^{2p}x\tan x dx=\int(\sec^{2}x-1)^{p}\tan x dx\end{align*}$
::tan2pxtanxdx(sec2x- 1)ptanxdx

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

$\begin{align*}\int(\sec^{2}x-1)^{p}\tan x dx=\int(u^{2}-1)^{p}\frac{du}{u}\end{align*}$
:sec2x -1)pduu

(4) Divide the polynomial function in $\begin{align*}u\end{align*}$ by $\begin{align*}u\end{align*}$ .
:4) u除以 u 的多元函数除以 u。

(5) Integrate the resulting terms of $\begin{align*}u\end{align*}$ .
:5) 将由此产生的u条款纳入其中。

(6) Substitute back $\begin{align*}\sec x\end{align*}$ for $\begin{align*}u\end{align*}$ .
:6) 以 u 取代 secx。

$\begin{align*}\tan^{2}x=\sec^{2}x-1\end{align*}$
::tan2\\\ x=sec2\\\\\\ x- 1

$\begin{align*}u=\sec x\end{align*}$ and $\begin{align*}du=\sec x \tan x dx\end{align*}$ , or $\begin{align*}\frac{du}{u}=\tan x dx\end{align*}$
::u=secx 和 du=secxtanxdx, 或 duu=tanxdx

$\begin{align*}m\end{align*}$ is even
::以米为偶数

$\begin{align*}(m=2p)\end{align*}$
:m=2p)

(1) Convert the integral:
:1) 转换元件:

$\begin{align*}\int\tan^{m}x dx=\int\tan^{2p}xdx\end{align*}$
::\ tanmxdxtan2pxdx

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

$\begin{align*}\int\tan^{2p}x dx=\int(\sec^{2}x-1)^{p} dx\end{align*}$
::\ tan2pxdx( sec2x- 1) pdx

(3) Expand the integrand in powers (even) of $\begin{align*}\sec x\end{align*}$ .
:3) 扩大cEZX的实权(甚至)范围。

(4) Integrate term by term the polynomial of even powers of $\begin{align*}\sec x\end{align*}$ using the reduction formula:
:4) 使用削减公式的cEx偶数权力的多等式综合术语:

$\begin{align*}\int\sec^{n}xdx=\frac{\sec^{n-2}x \tan x}{n-1}+\frac{n-2}{n-1}\int \sec^{n-2}xdx\end{align*}$
::@ secnxdx=secn-2xtanxn-1+n-2n- 1secn-2xdx

$\begin{align*}\tan^{2}x=(\sec^{2}x-1)\end{align*}$
::tan2_%x=( sec2_%x- 1)

Let's apply the formula to evaluate $\begin{align*}\int \tan^{3}xdx\end{align*}$
::让我们应用公式来评价tan3xdx

Approach I
::办法一

Reduction Formula

\begin{aligned} \int \tan^{m} x d x & = \frac{\tan^{m - 1} x}{m - 1} - \int \tan^{m - 2} x d x \\ \int \tan^{3} x d x & = \frac{\tan^{2} x}{2} - \int \tan x d x \\ \int \tan^{3} x d x & = \frac{\tan^{2} x}{2} + \ln | \sec x | + C \dots or = \frac{\tan^{2} x}{2} - \ln | \cos x | + C \end{aligned}

$\begin{align*}\int\tan^{m}xdx & = \frac{\tan^{m-1}x}{m-1}-\int\tan^{m-2}xdx\\ \int\tan^{3}xdx & =\frac{\tan^{2}x}{2}-\int\tan xdx\\ \int \tan^{3}xdx & =\frac{\tan^{2}x}{2}+\ln |\sec x|+C \qquad \ldots \text{or}=\frac{\tan^{2}x}{2}-\ln |\cos x|+C\end{align*}$
::减少公式xdx=tanm-1xm-1tanm-2xdx}}3xdx=tan2xxx}xn3xdx=tan2x2x2x2x2xxxxxx=lnsexxC...or=tan2x2_nx2_xxxx{C

Approach II
::方法二方法二

$\begin{align*}m\end{align*}$ odd

\begin{aligned} \int \tan^{3} x d x & = \int \tan^{2} x \tan x d x \\ = \int (\sec^{2} x - 1) \tan x d x \\ = \int (u^{2} - 1) \frac{d u}{u} \\ = \frac{u^{2}}{2} - \ln | u | + C \\ = \frac{\sec^{2} x}{2} - \ln | \sec x | + C \dots or \frac{\tan^{2} x}{2} - \ln | \sec x | + C^{'}, where \\ C^{'} = C - \frac{1}{2} \end{aligned}

$\begin{align*}\int \tan^{3}xdx & = \int \tan^{2}x\tan xdx\\ & = \int (\sec^{2}x-1)\tan x dx\\ & = \int(u^{2}-1)\frac{du}{u}\\ & = \frac{u^{2}}{2}- \ln |u|+C\\ & = \frac{\sec^{2}x}{2}-\ln|\sec x|+C \qquad \ldots \text{or} \ \frac{\tan^{2}x}{2}-\ln|\sec x|+C^\prime, \text{where}\\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad C^\prime=C-\frac{1}{2} \end{align*}$
::\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\C\\\\\\\\\\\\\\\\\\\\\\\\\\\\C\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Problem Form II: $\begin{align*}\int\sec^{n}xdx\end{align*}$
::问题表二:=========================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================

For the case of $\begin{align*}n=1\end{align*}$ , the integral is evaluated as follows:
::对于n=1的情况,整体体评估如下:

\begin{aligned} \int \sec x d x & = \int \sec x (\frac{\sec x + \tan x}{\sec x + \tan x}) d x \\ = \int \frac{\sec^{2} x + \sec x \tan x}{\sec x + \tan x} d x \\ = \int \frac{d u}{u} & \dots Using u -substitution: \\ u = \sec x + \tan x and \\ d u = (\sec x \tan x + \sec^{2} x) d x \\ = \ln | u | + C \\ \int \sec x d x & = \ln | \sec x + \tan x | + C \end{aligned}

$\begin{align*}\int\sec xdx & =\int\sec x\left(\frac{\sec x+ \tan x}{\sec x+ \tan x}\right)dx\\ & = \int\frac{\sec^{2}x+\sec x\tan x}{\sec x+\tan x}dx\\ & = \int \frac{du}{u} && \ldots \text{Using}\ u \text{-substitution:}\\ & && u=\sec x+\tan x \ \text{and}\\ & && du=(\sec x\tan x+\sec^{2}x)dx\\ & = \ln|u|+C\\ \int\sec xdx & = \ln|\sec x+\tan x|+C\end{align*}$
::@ sec_ xxxx @ sec_ xx+tan_ xxx}dx @ sec2}x+sec*xxxx*xxxx*duu...Using u- 替代: u=sec*x+tan*x和du=(sec*xx+sec2}dx=lnC*sec*xxxx=ln{sec*xxxxx=xxxx}

For evaluating higher powers of secant use the following reduction formula:
::在评价更高级的分离权时,使用以下削减公式:

Reduction Formula: $\begin{align*}\int\sec^{n}xdx\end{align*}$
::减少公式:secnxdx

\begin{aligned} \int \sec^{n} x d x = \frac{\sec^{n - 2} x \tan x}{n - 1} + \frac{n - 2}{n - 1} \int \sec^{n - 2} x d x \end{aligned}

$\begin{align*}\int\sec^{n}xdx=\frac{\sec^{n-2}x\tan x}{n-1}+\frac{n-2}{n-1}\int\sec^{n-2}xdx\end{align*}$
::@ secnxdx=secn-2xtanxn-1+n-2n- 1secn-2xdx

If $\begin{align*}n=2\end{align*}$ , use of the reduction formula yields $\begin{align*}\int\sec^{2}xdx\end{align*}$
::如果 n= 2, 则使用减排公式的收益率 sec2xdx

\begin{aligned} \int \sec^{2} x d x & = \frac{\sec^{2 - 2} x \tan x}{2 - 1} + \frac{2 - 2}{2 - 1} \int \sec^{2 - 2} x d x \\ = \tan x + C \end{aligned}

$\begin{align*}\int\sec^{2}xdx & = \frac{\sec^{2-2}x\tan x}{2-1}+\frac{2-2}{2-1}\int\sec^{2-2}xdx\\ & = \tan x+C\end{align*}$
::sec2xdx=sec2-2xtanx2-1+2-2-22-1sec2-2-2xdx=tanx+C

This is the familiar result.
::这是人们所熟知的结果。

Examples
::实例

Example 1
::例1

Earlier, you were asked about the identity relationship between the tangent and secant functions and their . Also, you were asked to evaluate $\begin{align*}\int\tan x\sec^{2}xdx\end{align*}$ ?
::早些时候,有人问及过切函数和断层函数之间的身份关系及其关系。此外,有人要求你评估 tanxsec2xdx?

The identity relationship is: $\begin{align*}\tan^{2}x+1=\sec^{2}x\end{align*}$ , which can be derived from $\begin{align*}\sin^{2}x+\cos^{2}x=1\end{align*}$ by dividing both sides of the equation by $\begin{align*}\cos^{2}x\end{align*}$ .
::身份关系是: tan2x+1=sec2x, 可以从 sin2x+cos2x=1 中得出, 其方法是将方程式的两边除以 cos2x 。

The derivative relationships are: $\begin{align*}\frac{d}{dx}\tan x=\sec^{2}x\end{align*}$ and $\begin{align*}\frac{d}{dx}\sec x=\sec x\tan x\end{align*}$ .
::衍生关系是:ddxtanx=sec2x和ddxseçx=secxtanxx。

To evaluate $\begin{align*}\int\tan x\sec^{2}xdx\end{align*}$ , let $\begin{align*}u=\tan x, du=\sec^{2}xdx\end{align*}$ .
::来评估 nxsec2xdx, let u=tanx, du=sec2xdx。

Then $\begin{align*}\int\tan x\sec^{2}xdx=\int u du=\frac{u^{2}}{2}+C=\frac{\tan^{2}x}{2}+C\end{align*}$ .
::然后是xsec2xdxudu=u22+C=tan2x2+C。

Example 2
::例2

Evaluate $\begin{align*}\int\sec^{3}xdx\end{align*}$ .
::评估 sec3xdx。

We use the reduction formula above with $\begin{align*}n=3\end{align*}$ .
::我们使用以 n=3 表示的上面的削减公式。

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

$\begin{align*}\int\sec^{n} xdx & = \frac{\sec^{n-2}x\tan x}{n-1}+\frac{n-2}{n-1}\int\sec^{n-2}xdx\\ \int\sec^{3}xdx & = \frac{\sec x\tan x}{2}+\frac{1}{2}\int\sec xdx\\ \int\sec^{3}xdx & = \frac{\sec x\tan x}{2}+\frac{1}{2}\ln|\sec x+\tan x|+C\end{align*}$
::@ secnxdx=secn- 2xtanxn- 1+n- 2n- 1secn- 2xdx}xdx=secxtanxx2+12secxdx}secxxxx}xdx3xdx=secxtanx2+12ln{sec}xxx

Problem Form III: $\begin{align*}\int\tan^{m}x \sec^{n}xdx\end{align*}$ , $\begin{align*}m\ge 1\end{align*}$ and $\begin{align*}n\ge 1\end{align*}$
::问题三:% tanmxsecnxdx, 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。

$\begin{align}\int\tan^{m}x \sec^{n}xdx\end{align}$	Procedure ::程序程序程序程序程序程序程序	Key Identities ::关键特征
$\begin{align}m\end{align}$ is odd ::m是奇数,m是奇数 $\begin{align}(m=2p+1)\end{align}$ :m=2p+1)	(1) Convert integrand by splitting off a $\begin{align}\tan x\end{align}$ and $\begin{align}\sec x\end{align}$ : :1) 通过分割一个 tanx 和 sex 转换整数 : $\begin{align}\int\tan^{m}x\sec^{n}xdx=\int\tan^{2p}x\sec^{n-1}x\sec x \tan xdx\end{align}$ ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}但愿如此 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080} (2) Convert the integrand to all $\begin{align}\sec x\end{align}$ : :2) 将整数转换成全部秒==x: $\begin{align}\int(\sec^{2}x-1)^p \sec^{n-1}x\sec x \tan x dx\end{align}$ :sec2x-1) psecn - 1xsecxtanxdx (3) Use $\begin{align}u\end{align}$ -substitution to change variables: :3) 使用 u 替代来改变变量: $\begin{align}\int(\sec^{2}x-1)^{p}\sec^{n-1}x\sec x\tan xdx=\int(u^{2}-1)^{p}u^{n-1}du\end{align}$ : sec2_ x- 1) psecn - 1 xsec xtanxdx( u2 - 1) pun - 1du (4) Expand the integrand to a polynomial in $\begin{align}u\end{align}$ . :4) 将原数扩大至在u的多元体。 (5) Integrate the resulting polynomial in $\begin{align}u\end{align}$ . :5) 将由此产生的多元性并入u。 (6) Substitute back $\begin{align}\sec x\end{align}$ for $\begin{align}u\end{align}$ . :6) 以 u 取代 secx。	$\begin{align}\tan^{2}x=\sec^{2}x-1\end{align}$ ::tan2\\\ x=sec2\\\\\\ x- 1 $\begin{align}u=\sec x\end{align}$ and $\begin{align}du=\sec x\tan xdx\end{align}$ ::u=secx 和 du=secxtanxdx
$\begin{align}n\end{align}$ is even $\begin{align}(n=2p)\end{align}$ ::n 是偶数 (n=2p)	(1) Convert integrand by splitting off a $\begin{align}\sec^{2}x\end{align}$ : :1) 分秒转换整数,转换成秒2x: $\begin{align}\int\tan^{m}x\sec^{n}xdx=\int\tan^{m}x\sec^{2p-2}x\sec^{2}xdx\end{align}$ :: (2) Convert $\begin{align}\sec^{2p-2}x\end{align}$ to powers of $\begin{align}(\tan^{2}x+1)\end{align}$ : :2) 将秒2p-2x 转换为 (tan2x+1) 的功率 : $\begin{align}\int\tan^{m}x\sec^{n}xdx=\int\tan^{m}x(\tan^{2}x+1)^{2p-2}x\sec^{2}xdx\end{align}$ :tan2x+1) 2p- 2xsec2xdx (3) Use $\begin{align}u\end{align}$ -substitution to change variables: :3) 使用 u 替代来改变变量: $\begin{align}\int\tan^{m}x(\tan^{2}x+1)^{2p-2}x\sec^{2}xdx=\int u^{m}(u^{2}+1)^{2(p-1)}du\end{align}$ ::* tanmx( tan2x+1) 2p- 2xsec2xdxum( u2+1) 2( p- 1) du (4) Expand the integrand to form a polynomial in $\begin{align}u\end{align}$ . :4) 在u中扩大成一个多面体。 (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 $\begin{align}\tan x\end{align}$ for $\begin{align}u\end{align}$ . :6) 以 u 代替 tanx 。	$\begin{align}\sec^{2}x=\tan^{2}x+1\end{align}$ ::秒2x=tan2x+1 $\begin{align}u=\tan x\end{align}$ and $\begin{align}du=\sec^{2}xdx\end{align}$ ::u=tanx 和 du=sec2xdx
$\begin{align}m\end{align}$ is even $\begin{align}(m=2p)\end{align}$ and $\begin{align}n\end{align}$ is odd. ::m 等于 (m=2p) , n 是奇数。	(1) Convert the integral: :1) 转换元件: $\begin{align}\int\tan^{m}x\sec^{n}xdx=\int\tan^{2p}x\sec^{n}xdx\end{align}$ ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}但愿如此 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不胜感激 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080} (2) Use the identity to eliminate powers of $\begin{align}\tan x\end{align}$ : :2) 使用身份来消除Tanx的权力: $\begin{align}\int\tan^{2p}x\sec^{n}xdx=\int(\sec^{2}x-1)^{p}\sec^{n}xdx\end{align}$ ::\ tan2pxsecnxdx( sec2x- 1) psecnxdx (3) Expand the integrand to get powers of $\begin{align}\sec x\end{align}$ . : 三 ) 扩大原数 , 以获得秒 x 的功率。 (4) Integrate the integrand term by term using the reduction formula: :4) 使用削减公式将整数术语按术语合并: $\begin{align}\int\sec^{n}xdx=\frac{\sec^{n-2}x\tan x}{n-1}+\frac{n-2}{n-1}\int\sec^{n-2}xdx\end{align}$ ::@ secnxdx=secn-2xtanxn-1+n-2n- 1secn-2xdx	$\begin{align}\tan^{2}x=\sec^{2}x-1\end{align}$ ::tan2\\\ x=sec2\\\\\\ x- 1

Example 3
::例3

Evaluate $\begin{align*}\int\tan^{2}x\sec^{4}xdx\end{align*}$ .
::评估#tan2xsec4xdx。

Here $\begin{align*}n=4\end{align*}$ , is even so the procedure is as follows:
::这里n=4,甚至程序如下:

\begin{aligned} \int \tan^{2} x \sec^{4} x d x & = \int \tan^{2} x \sec^{2} \sec^{2} x d x \dots Split off \sec^{2} x \\ = \int \tan^{2} x (\tan^{2} x + 1) \sec^{2} x d x \\ = \int u^{2} (u^{2} + 1) d u \dots Use u -subtitution: \\ u = \tan x and d u = \sec^{2} x d x \\ = \int (u^{4} + u^{2}) d u \\ = \frac{u^{5}}{5} + \frac{u^{3}}{3} + C \\ \int \tan^{2} x \sec^{4} x d x & = \frac{\tan^{5} x}{5} + \frac{\tan^{3} x}{3} + C \dots Substitute \tan x for u \end{aligned}

$\begin{align*}\int\tan^{2}x\sec^{4}xdx & = \int\tan^{2}x\sec^{2}\sec^{2}xdx \qquad\qquad \ldots \text{Split off} \ \sec^{2}x\\ & = \int\tan^{2}x(\tan^{2}x+1)\sec^{2}xdx\\ & = \int u^{2}(u^{2}+1)du \quad\quad\qquad\qquad\ldots\text{Use}\ u \text{-subtitution:}\\ & \qquad\qquad\quad\qquad\qquad\qquad\qquad\qquad\qquad u=\tan x \text{ and } du=\sec^{2}xdx\\ & = \int(u^{4}+u^{2})du\\ & = \frac{u^{5}}{5}+\frac{u^{3}}{3}+C\\ \int\tan^{2}x\sec^{4}xdx & =\frac{\tan^{5}x}{5}+\frac{\tan^{3}x}{3}+C \qquad\qquad \ldots \text{Substitute}\tan x \ \text{for} \ u\end{align*}$
::tan2xsec4xdx}xse2xese2xxxxx... splish sec2xx}xxxxxxxxxxu2(u2+1du)...使用 u-sutututution: u=tan*xx 和 du= u+u+u33+C2xsec4}xdxx=tan5}x5+tan3x3+C...suittutanx for u= u=tan*xxxxxxxxxxx=u=55+u33+C2xsec4}xdxxxxxxx=tan5}x5+tan3x3+C...suittutanxx for u

Example 4
::例4

Evaluate $\begin{align*}\int\tan^{3}x\sec^{3}xdx\end{align*}$ .
::评估 3 xsec3 xdx。

Here $\begin{align*}m=3\end{align*}$ , and we use the following procedure:
::这里m=3,我们使用以下程序:

\begin{aligned} \int \tan^{3} x \sec^{3} x d x & = \int \tan^{2} x \sec^{2} x \sec x \tan x d x \\ = \int (\sec^{2} x - 1) \sec^{2} x \sec x \tan x d x \\ = \int (u^{2} - 1) u^{2} d u & \dots Using u -substitution: \\ u = \sec x and d u = \sec x \tan x d x \\ = \int (u^{4} - u^{2}) d u \\ = \frac{u^{5}}{5} - \frac{u^{3}}{3} + C \\ \int \tan^{2} x \sec^{4} x d x & = \frac{\sec^{5} x}{5} - \frac{\sec^{3} x}{3} + C & \dots Substitute \sec x for u \end{aligned}

$\begin{align*}\int\tan^{3}x\sec^{3}xdx & =\int\tan^{2}x\sec^{2}x \sec x \tan x dx\\ & = \int(\sec^{2}x-1)\sec^{2}x\sec x\tan xdx\\ & = \int(u^{2}-1)u^{2}du && \ldots\text{Using}\ u \text{-substitution:}\\ & && u=\sec x \text{ and } du=\sec x\tan xdx\\ & = \int(u^{4}-u^{2})du\\ & = \frac{u^{5}}{5}-\frac{u^{3}}{3}+C\\ \int\tan^{2}x\sec^{4}xdx & =\frac{\sec^{5}x}{5}-\frac{\sec^{3}x}{3}+C && \ldots\text{Substitute }\sec x \ \text{for} \ u\end{align*}$
::

Review
::回顾

For #1-7, evaluate the integrals.
::为了1 -7, 评估整体体。

$\begin{align*}\int\tan^{4}xdx\end{align*}$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}但... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}但... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

$\begin{align*}\int\tan^{5}xdx\end{align*}$
::\ tan5xdx

$\begin{align*}\int\sec^{4}xdx\end{align*}$
::sec4xdx

$\begin{align*}\int\sec^{5}xdx\end{align*}$
::sec5xdx

$\begin{align*}\int\sec^{4}x\tan^{3}xdx\end{align*}$
::@ sec4xtan3xdx

$\begin{align*}\int\tan^{4}x\sec xdx\end{align*}$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}但愿如此 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

$\begin{align*}\int\sqrt{\tan x}\sec^{4}xdx\end{align*}$
::{\fn方正准圆简体\fs12\1cHC9ECC4\b1\bord1\shad1\3cH2F2F2F} xsec4}xdxxxxxxx

Graph and then find the volume of the solid that results when the region enclosed by $\begin{align*}y=\sec x\tan x,y=0,y=\sqrt{2},x=0\end{align*}$ , and $\begin{align*}x=\pi\end{align*}$ is revolved around the $\begin{align*}x\end{align*}$ -axis.
::图形然后找到以 y=secxtanx,y=0,y=2,y=2,x=0和 x围绕 x 轴旋转的区域所得出的固体体积。

For #9-15, evaluate the integrals.
::915,评估整体体。

$\begin{align*}\int x\sec^{2}(3x^{2})dx\end{align*}$
::xsec2( 3x2) dx

$\begin{align*}\int x^{2}\tan(4x^{3})\sec^{2}(4x^{3})dx\end{align*}$
::x2tan( 4x3) sec2( 4x3) dx

$\begin{align*}\int \sec^{7}(x)\tan^{5}(x)dx\end{align*}$
::sec7(x)tan5(x)dx

$\begin{align*}\int\sec^{4}(x)\tan^{7}(x)dx\end{align*}$
::sec4(x)tan7(x)dx

$\begin{align*}\int\sec^{3}(x)\tan^{3}(x)dx\end{align*}$
:x)tan3(x)dx

$\begin{align*}\int\sec^{4}(x)\tan^{2}(x)dx\end{align*}$
:x)tan2(x)dx

$\begin{align*}\int\sec^{5}(x)\tan(x)dx\end{align*}$
::sec5(x)tan(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。

$\begin{align}\int\tan^{m}x \sec^{n}xdx\end{align}$	Procedure ::程序程序程序程序程序程序程序	Key Identities ::关键特征
$\begin{align}m\end{align}$ is odd ::m是奇数,m是奇数 $\begin{align}(m=2p+1)\end{align}$ :m=2p+1)	(1) Convert integrand by splitting off a $\begin{align}\tan x\end{align}$ and $\begin{align}\sec x\end{align}$ : :1) 通过分割一个 tanx 和 sex 转换整数 : $\begin{align}\int\tan^{m}x\sec^{n}xdx=\int\tan^{2p}x\sec^{n-1}x\sec x \tan xdx\end{align}$ ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}但愿如此 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080} (2) Convert the integrand to all $\begin{align}\sec x\end{align}$ : :2) 将整数转换成全部秒==x: $\begin{align}\int(\sec^{2}x-1)^p \sec^{n-1}x\sec x \tan x dx\end{align}$ :sec2x-1) psecn - 1xsecxtanxdx (3) Use $\begin{align}u\end{align}$ -substitution to change variables: :3) 使用 u 替代来改变变量: $\begin{align}\int(\sec^{2}x-1)^{p}\sec^{n-1}x\sec x\tan xdx=\int(u^{2}-1)^{p}u^{n-1}du\end{align}$ : sec2_ x- 1) psecn - 1 xsec xtanxdx( u2 - 1) pun - 1du (4) Expand the integrand to a polynomial in $\begin{align}u\end{align}$ . :4) 将原数扩大至在u的多元体。 (5) Integrate the resulting polynomial in $\begin{align}u\end{align}$ . :5) 将由此产生的多元性并入u。 (6) Substitute back $\begin{align}\sec x\end{align}$ for $\begin{align}u\end{align}$ . :6) 以 u 取代 secx。	$\begin{align}\tan^{2}x=\sec^{2}x-1\end{align}$ ::tan2\\\ x=sec2\\\\\\ x- 1 $\begin{align}u=\sec x\end{align}$ and $\begin{align}du=\sec x\tan xdx\end{align}$ ::u=secx 和 du=secxtanxdx
$\begin{align}n\end{align}$ is even $\begin{align}(n=2p)\end{align}$ ::n 是偶数 (n=2p)	(1) Convert integrand by splitting off a $\begin{align}\sec^{2}x\end{align}$ : :1) 分秒转换整数,转换成秒2x: $\begin{align}\int\tan^{m}x\sec^{n}xdx=\int\tan^{m}x\sec^{2p-2}x\sec^{2}xdx\end{align}$ :: (2) Convert $\begin{align}\sec^{2p-2}x\end{align}$ to powers of $\begin{align}(\tan^{2}x+1)\end{align}$ : :2) 将秒2p-2x 转换为 (tan2x+1) 的功率 : $\begin{align}\int\tan^{m}x\sec^{n}xdx=\int\tan^{m}x(\tan^{2}x+1)^{2p-2}x\sec^{2}xdx\end{align}$ :tan2x+1) 2p- 2xsec2xdx (3) Use $\begin{align}u\end{align}$ -substitution to change variables: :3) 使用 u 替代来改变变量: $\begin{align}\int\tan^{m}x(\tan^{2}x+1)^{2p-2}x\sec^{2}xdx=\int u^{m}(u^{2}+1)^{2(p-1)}du\end{align}$ ::* tanmx( tan2x+1) 2p- 2xsec2xdxum( u2+1) 2( p- 1) du (4) Expand the integrand to form a polynomial in $\begin{align}u\end{align}$ . :4) 在u中扩大成一个多面体。 (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 $\begin{align}\tan x\end{align}$ for $\begin{align}u\end{align}$ . :6) 以 u 代替 tanx 。	$\begin{align}\sec^{2}x=\tan^{2}x+1\end{align}$ ::秒2x=tan2x+1 $\begin{align}u=\tan x\end{align}$ and $\begin{align}du=\sec^{2}xdx\end{align}$ ::u=tanx 和 du=sec2xdx
$\begin{align}m\end{align}$ is even $\begin{align}(m=2p)\end{align}$ and $\begin{align}n\end{align}$ is odd. ::m 等于 (m=2p) , n 是奇数。	(1) Convert the integral: :1) 转换元件: $\begin{align}\int\tan^{m}x\sec^{n}xdx=\int\tan^{2p}x\sec^{n}xdx\end{align}$ ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}但愿如此 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不胜感激 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080} (2) Use the identity to eliminate powers of $\begin{align}\tan x\end{align}$ : :2) 使用身份来消除Tanx的权力: $\begin{align}\int\tan^{2p}x\sec^{n}xdx=\int(\sec^{2}x-1)^{p}\sec^{n}xdx\end{align}$ ::\ tan2pxsecnxdx( sec2x- 1) psecnxdx (3) Expand the integrand to get powers of $\begin{align}\sec x\end{align}$ . : 三 ) 扩大原数 , 以获得秒 x 的功率。 (4) Integrate the integrand term by term using the reduction formula: :4) 使用削减公式将整数术语按术语合并: $\begin{align}\int\sec^{n}xdx=\frac{\sec^{n-2}x\tan x}{n-1}+\frac{n-2}{n-1}\int\sec^{n-2}xdx\end{align}$ ::@ secnxdx=secn-2xtanxn-1+n-2n- 1secn-2xdx	$\begin{align}\tan^{2}x=\sec^{2}x-1\end{align}$ ::tan2\\\ x=sec2\\\\\\ x- 1

Section outline

General

将占地人和唐恩特人的权力结合起来

Integrating Powers of Secants and Tangents ::将占地人和唐恩特人的权力结合起来

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)