7.12 涉及反反双曲函数的衍生物和综合物

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  涉及反反双曲函数的衍生物和综合物

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  In the previous concept that introduced inverse hyperbolic functions , the concept question asked the name of the curve that described the path of a weight attached to a taut rope of a given length being pulled by someone in a direction perpendicular to the initial position of the weight. The curve is defined by the equation $\begin{align*}y=a \ \text{sech}^{-1}\left(\frac{x}{a}\right)-\sqrt{a^{2}-x^{2}}\end{align*}$ , and is called a tractrix . It can also describe the path taken by the middle of the rear tire axle of a tractor trailer as it makes a turn perpendicular to its original direction of motion.
  ::在前一个引入反双曲函数的概念中,概念问题询问曲线的名称,其中描述了某人在与重量初始位置相垂直的方向上拉动某一长度的拖绳所附加的重量线的路径,曲线由公式y=a sech-1(xa) a2-x2定义,并称为轮轴,还可以描述拖拉机拖车后胎轴中间的路径,因为它使旋转与最初运动方向垂直。

  

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  Given the tractrix described in the previous concept by $\begin{align*}y=10 \ \text{sech}^{-1}\left(\frac{x}{10}\right)-\sqrt{10^{2}-x^{2}}\end{align*}$ , can you determine the actual distance traveled by the weight along the curve between  $\begin{align*}x=10 \ ft\end{align*}$ (the start) and  $\begin{align*}x=2 \ ft\end{align*}$ (the end)?
  ::根据y=10 sech- 1(x10) 102- x2 所描述的上一个概念中的轮廓,您能否确定 x= 10 ft (起始时间) 和 x= 2 ft (结尾时间) 之间曲线的重量所穿行的实际距离 ?

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  Derivatives and Integrals with Inverse Hyperbolic Functions
  ::具有反反双曲函数的衍生物和综合物

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  Recall from the previous concepts that the hyperbolic functions and their inverses are defined as:
  ::回顾以前的概念,双曲函数及其反作用被定义为:

  Table 1. Summary of Hyperbolic Functions and their Inverses

  	Inverse Hyperbolic Function	Domain	Range
  $\begin{align*}\sinh x=\frac{e^{x}-e^{-x}}{2}\end{align*}$	$\begin{align*}\sinh^{-1} x=\ln \left ( x+\sqrt{x^{2}+1} \right )\end{align*}$	$\begin{align*}-\infty\le x \le \infty\end{align*}$	$\begin{align*}(-\infty,\infty)\end{align*}$
  $\begin{align*}\cosh x=\frac{e^{x}+e^{-x}}{2}\end{align*}$	$\begin{align*}\cosh^{-1} x=\ln \left ( x+\sqrt{x^{2}-1} \right )\end{align*}$	$\begin{align*}x\ge 1\end{align*}$	$\begin{align*}[0,\infty)\end{align*}$
  $\begin{align*}\tanh x=\frac{\sinh x}{\cosh x}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\end{align*}$	$\begin{align*}\tanh^{-1}x=\frac{1}{2}\ln\frac{1+x}{1-x}\end{align*}$	$\begin{align*}|x|<1\end{align*}$	$\begin{align*}(-\infty,\infty)\end{align*}$
  $\begin{align*}\text{csch} \ x=\frac{1}{\sinh x} =\frac{2}{e^{x}-e^{-x}}\end{align*}$	播放段落 $\begin{align*}\text{csch}^{-1} x =\sinh^{-1}\left(\frac{1}{x}\right)\end{align*}$ ::csch- 1x=sinh-1(1x) 播放段落 $\begin{align*}=\ln\left(\frac{1}{x}+\frac{\sqrt{1+x^{2}}}{|x|}\right)\end{align*}$ ::=n( 1x1+x2x)	$\begin{align*}x\ne 0\end{align*}$	$\begin{align*}(-\infty,\infty)\end{align*}$
  $\begin{align*}\text{sech} \ x=\frac{1}{\cosh x}=\frac{2}{e^{x}+e^{-x}}\end{align*}$	播放段落 $\begin{align*}\text{sech}^{-1} x=\coth^{-1}\left(\frac{1}{x}\right)\end{align*}$ ::sech- 1x=coth-1( 1x) 播放段落 $\begin{align*}=\ln\left(\frac{1+\sqrt{1-x^{2}}}{x}\right)\end{align*}$ ::=n(11-x2x)	$\begin{align*}0< x\le 1\end{align*}$
  $\begin{align*}\coth x=\frac{1}{\tanh x}=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\end{align*}$	播放段落 $\begin{align*}\coth^{-1}x=\tanh^{-1}\left(\frac{1}{x}\right)\end{align*}$ ::COth- 1x=tanh- 1(1x) 播放段落 $\begin{align*}=\frac{1}{2}\ln\frac{x+1}{x-1}\end{align*}$ ::= 12lnx+1x-1	$\begin{align*}|x|>1\end{align*}$

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  The and integrals follow easily from the above.
  ::从上述情况看,整体和整体都很容易。

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  Derivatives of the Inverse Hyperbolic Functions
  ::反双曲函数的衍生因素

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  Finding the derivative of each of the inverse hyperbolic functions is just a matter of differentiating each of the above expressions. If we let the argument of each inverse hyperbolic function be $\begin{align*}u(x)\end{align*}$ , then the generalized derivatives of the inverse hyperbolic functions are:
  ::查找每个反双曲函数的衍生物只是区别上述每个表达式的问题。如果我们让每个反双曲函数的参数为 u(x),那么反双曲函数的普遍衍生物是:

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  Let  $\begin{align*}u(x)\end{align*}$ be a differentiable function
  ::Letu(x) 是一个可区别的函数 Letu( x)

  Table 2. Derivatives of the Inverse Hyperbolic Functions

  $\begin{align*}\frac{d}{dx}\sinh^{-1}u=\frac{1}{\sqrt{1+u^{2}}}\frac{du}{dx} \ \quad u\in \mathbb{R}\end{align*}$	$\begin{align*}\frac{d}{dx}\text{csch}^{-1} \ u=\frac{-1}{|u|\sqrt{1+u^{2}}}\frac{du}{dx} \ \quad u\ne 0\end{align*}$
  $\begin{align*}\frac{d}{dx}\cosh^{-1} u=\frac{1}{\sqrt{u^{2}-1}}\frac{du}{dx} \ \quad u>1\end{align*}$	$\begin{align*}\frac{d}{dx}\text{sech}^{-1} \ u=\frac{-1}{u\sqrt{1-u^{2}}}\frac{du}{dx} \ \ \quad 0 < u < 1\end{align*}$
  $\begin{align*}\frac{d}{dx}\tanh^{-1} u=\frac{1}{1-u^{2}}\frac{du}{dx} \ \ \ \qquad |u|<1\end{align*}$	$\begin{align*}\frac{d}{dx}\coth^{-1} u=\frac{1}{1-u^{2}}\frac{du}{dx} \ \ \quad \qquad |u|>1\end{align*}$

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  To see how easily the derivatives are derived, look at the following example.
  ::为了解衍生物的产生多么容易,请以下列例子为例。

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  Derive the derivative of $\begin{align*}\sinh^{-1}x\end{align*}$ :
  ::产生sinh - 1x的衍生物:

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  We know that $\begin{align*}\sinh^{-1} x=\ln \left (x+\sqrt{x^{2}+1} \right )\end{align*}$ . Let $\begin{align*}u=x+\sqrt{x^{2}+1}\end{align*}$ ; then $\begin{align*}\sinh^{-1} x=\ln(u)\end{align*}$ .
  ::我们知道 sinh- 1x=ln( xx2+1) 。 Let u=xx2+1; 然后sinh- 1x=ln(u) 。

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  Evaluating the derivative results in these steps:
  ::评估衍生产品后,将采取以下步骤:

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  $$\begin{align*}\frac{d}{dx}\sinh^{-1} \ x & = \frac{d}{dx}\ln(u)\\ & = \frac{1}{u}\frac{du}{dx}\\ & = \frac{1}{x+\sqrt{x^{2}+1}}\cdot \left(1+\frac{1}{2\sqrt{x^{2}+1}}\cdot 2x\right)\\ & = \frac{1}{x+\sqrt{x^{2}+1}}\cdot \left(\frac{x+\sqrt{x^{2}+1}}{\sqrt{x^{2}+1}}\right)\\ & = \frac{1}{\sqrt{x^{2}+1}}\\\end{align*}$$
  ::ddxsinh - 1 x=ddxln( u) =1ududdx=1xxxxx2x2+1}( 1+12_x2+1_2x) =1xxxxx2+1x2+1xx2+1}( xx2+1x2+1) =1x2+1

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  Therefore, $\begin{align*}\frac{d}{dx}\sinh^{-1} x=\frac{1}{\sqrt{1+x^{2}}}\end{align*}$
  ::因此, ddxsinh - 1x=11+x2

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  Integrals Involving Inverse Hyperbolic Functions
  ::包含反反双曲函数的元件

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  Each of the derivative formulas presented above can be associated with an integral equation. For example,
  ::上述每种衍生公式都可以与一个整体等式相联系。

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  $$\begin{align*}\frac{d}{dx}[ar \sinh x]=\frac{1}{\sqrt{1+x^{2}}}\Leftrightarrow\int d[ar \sinh x]=\int \frac{1}{\sqrt{1+x^{2}}}dx=ar \sinh x+C.\end{align*}$$
  ::ddx[arsinhx]=11+x2************************************************ *************************************************************************************************************************************************************************************************************************************************************************************************************************************************************************************************************************************************************

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  Applying this procedure to the derivative of each inverse hyperbolic function results in these relationships:
  ::对每个反双曲函数的衍生物应用此程序导致这些关系:

  Table 3. Integrals Involving Inverse Hyperbolic functions: Basic Form

  播放段落 $\begin{align*}\int \frac{1}{\sqrt{1+x^{2}}}dx=\sinh^{-1}x+C\end{align*}$ ::11+x2dx=sinh-1x+C 播放段落 $\begin{align*}=\ln \left ( x+\sqrt{x^{2}+1} \right )+C\end{align*}$ ::=ln(xxx2+1)+C	$\begin{align*}\int \frac{1}{x\sqrt{1+x^{2}}}dx=-\text{csch}^{-1} x+C, \ \ \quad x\ne 0\end{align*}$
  $\begin{align*}\int \frac{1}{x^{2}-1}dx=\cosh^{-1} x+C, \quad x> 1\end{align*}$	$\begin{align*}\int \frac{1}{x\sqrt{1-x^{2}}}dx=-\text{sech}^{-1} x+C, \ \quad 0 < |x| < 1\end{align*}$
  $\begin{align*}\int \frac{1}{1-x^{2}}dx=\tanh^{-1} x+C, \quad |x|<1\end{align*}$	$\begin{align*}\int \frac{1}{1-x^{2}}dx=\coth^{-1} x+C, \ \ \ \qquad \quad |x|>1\end{align*}$

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  Notice that there are 5 different integrand forms. These integrand forms can all be generalized to provide a larger set of integrals that can be expressed as inverse hyperbolic functions. All of these generalized integrands can be setup so that using the $\begin{align*}u\end{align*}$ -substitution  $\begin{align*}u=\frac{b}{a}x\end{align*}$ transforms the integrand to one of the basic forms above. Three of the generalized integrands are shown below:
  ::注意有5种不同的整数形式。 这些整数形式都可以被普遍化, 以提供更大的组合, 可以用反双曲函数表示。 所有这些通用整数形式都可以设置, 以便使用 u =bax 将整数转换成上述基本形式之一。 其中三种通用整数如下:

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  $$\begin{align*}\frac{dx}{\sqrt{1+x^{2}}} & \Rightarrow \frac{A dx}{\sqrt{a^{2}+b^{2}x^{2}}}=\frac{A}{a}\frac{dx}{\sqrt{1+\left(\frac{b}{a}x\right)^2}}=\frac{A}{a}\frac{a}{b}\frac{du}{\sqrt{1+u^{2}}}\quad\qquad\qquad \ \ =\frac{A}{b}\cdot\frac{du}{\sqrt{1+u^{2}}}\\ \frac{dx}{1-x^{2}} & \Rightarrow \frac{A}{a^{2}-b^{2}x^{2}}dx=\frac{A}{a^{2}}\frac{1}{1-\left(\frac{b}{a}x\right)^2}dx=\frac{A}{a^{2}}\cdot \frac{a}{b}\frac{du}{1-u^{2}} \ \ \ \qquad\qquad \ =\frac{A}{ab}\cdot\frac{du}{1-u^{2}}\\ \frac{dx}{|x|\sqrt{1+x^{2}}} & \Rightarrow \frac{Adx}{|bx|\sqrt{a^{2}+b^{2}x^{2}}}=\frac{Adx}{\Big |\frac{b}{a}x \Big |a^{2}\sqrt{1+\left(\frac{b}{a}x\right)^2}}=\frac{Ab}{a^{2}}\frac{a}{b}\frac{du}{|u|\sqrt{1+u^{2}}}\quad=\frac{A}{a}\cdot\frac{du}{|u|\sqrt{1+u^{2}}}\end{align*}$$
  ::dx1+x2a2+x2=Aadx1+(ax)2=Aabdu1=Aabdu1+u2=Abdu1+u2x1-x2Aa2-b2x2dx=Aa211-(bax)2dx=Aa2}2dx=Aa2a2+b2x2=Aab}1+x2*1+x2}Adx2=Aadx1+(ax)1+b2+(abx)2=Adx_abdx1+2x1+(bax2)2=Aba2du1+u2=Aadu1+u2=Aaddu_1+u2

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  where $\begin{align*}A,a,\end{align*}$  and $\begin{align*}b\end{align*}$  are constants. The other 2 forms can be generalized in a similar fashion.
  ::其中A、a和b为常数。其他两种形式也可以以类似的方式加以普及。

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  The integrals of this generalized set of integrands can therefore also be expressed in terms of the inverse hyperbolic functions as presented in the table below:
  ::因此,还可用下表所示的反双曲函数来表示这组通用的数位数的综合体:

  $\begin{align*}A,a,\end{align*}$

  Table 4. Integrals Involving the Inverse Hyperbolic Functions: General Form

  播放段落 $\begin{align*}\int \frac{Adx}{\sqrt{a^{2}+b^{2}x^{2}}}=\frac{A}{b}\sinh^{-1}\left(\frac{b}{a}x\right)+C\end{align*}$ ::A2+b2x2=Absinh-1(轴)+C 播放段落 $\begin{align*}=\frac{A}{b}\ln\left(\frac{b}{a}x+\sqrt{\left(\frac{b}{a}\right)^{2}x^{2}+1}\right)+C\end{align*}$ ::=Abln( bax(ba) 2x2+1)+C	播放段落 $\begin{align*}\int\frac{Adx}{x\sqrt{a^{2}+b^{2}x^{2}}}=-\frac{A}{a}\text{csch}^{-1}\Big | \frac{b}{a}x \Big |+C\end{align*}$ ::* Adxxa2+b2x2*Aacsch-1baxC 播放段落 $\begin{align*}= - \frac{A}{a}\ln\left(\frac{1+\sqrt{1+\left(\frac{b}{a}\right)^{2}x^{2}}}{ \Big | \frac{b}{a}x \Big |}\right)+C \qquad x\ne 0\end{align*}$ ::======================================================================================================================================================
  播放段落 $\begin{align*}\int \frac{Adx}{\sqrt{b^{2}x^{2}-a^{2}}}=\frac{A}{b}\cosh^{-1}\left(\frac{b}{a}x\right)+C\end{align*}$ ::*Adxb2x2-a2=Acosh-1(轴)+C 播放段落 $\begin{align*}=\frac{A}{b}\ln\left(\frac{b}{a}x+\sqrt{\left(\frac{b}{a}\right)^{2}x^{2}+1}\right)+C\end{align*}$ ::=Abln( bax(ba) 2x2+1)+C	播放段落 $\begin{align*}\int\frac{Adx}{x\sqrt{a^{2}-b^{2}x^{2}}}=-\frac{A}{a}\text{sech}^{-1} \Big | \frac{b}{a}x \Big |+C\end{align*}$ ::* Adxxa2 -b2x2 *Aasaech -1bax*C *Adxxa2 -b2x2 *Aasaech -1*bax*C 播放段落 $\begin{align*}= - \frac{A}{a}\ln\left(\frac{1+\sqrt{1-\left(\frac{b}{a}\right)^{2}x^{2}}}{ \Big | \frac{b}{a}x \Big |}\right)+C \ \ 0 < |x| < 1\end{align*}$ ::=====================================================================================================================================================
  播放段落 $\begin{align*}\int \frac{Adx}{a^{2}-b^{2}x^{2}}=\frac{A}{ab}\tanh^{-1}\left(\frac{b}{a}x\right)+C, \ |x| < a\end{align*}$ ::*Adxa2-b2x2=Aabtanh-1(bax)+C,xa 播放段落 $\begin{align*}=\frac{A}{2ab}\ln\bigg|\frac{1+\frac{b}{a}x}{1-\frac{b}{a}x}\bigg|+C, \quad |x|\ne \frac{a}{b}\end{align*}$ ::= A2abln1+bax1 -baxC,xab	播放段落 $\begin{align*}\int \frac{Adx}{\sqrt{a^{2}-b^{2}x^{2}}}=\frac{A}{ab}\coth^{-1}\left(\frac{b}{a}x\right)+C, \ |x| > a\end{align*}$ ::*Adxa2 -b2x2=Aabcoth-1(bax)+C,xa 播放段落 $\begin{align*}=\frac{A}{2ab}\ln\bigg|\frac{1+\frac{b}{a}x}{1-\frac{b}{a}x}\bigg|+C, \quad |x|\ne \frac{a}{b}\end{align*}$ ::= A2abln1+bax1 -baxC,xab

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  Let's use the table above to evaluate the integral  $\begin{align*}\int\limits_{1}^{6}\frac{3}{\sqrt{9x^{2}-4}}dx\end{align*}$ .
  ::让我们用上表来评估 6139x2 -4dx。

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  The integral is of the form $\begin{align*}\int\frac{Adx}{\sqrt{b^{2}x^{2}-a^{2}}}\end{align*}$ , which has the general solution $\begin{align*}\frac{A}{b}\cosh^{-1}\left(\frac{b}{a}x \right)+C\end{align*}$ .
  ::有机体为表Adx_b2x2-2-a2,具有一般解决办法Abcosh-1(轴)+C。

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  Therefore $\begin{align*}\int\limits_1^6\frac{3}{\sqrt{9x^{2}-4}}dx=\frac{3}{3} \Big [ \cosh^{-1}\left(\frac{3}{2}x\right)\Big ]_1^6\end{align*}$ , and substituting the natural logarithm form yields

  $$\begin{align*}\int\limits_1^6\frac{3}{\sqrt{9x^{2}-4}}dx & =\ln\left(\frac{3}{2}6+\sqrt{\left(\frac{3}{2}\right)^{2}6^{2}+1}\right)-\ln\left(\frac{3}{2}1+\sqrt{\left(\frac{3}{2}\right)^{2}1^{2}+1}\right)\\ &=\ln \left ( 9+\sqrt{82} \right )-\ln\left(\frac{3}{2}+\sqrt{\frac{13}{4}}\right)\\ & = \ln\left(\frac{9+\sqrt{82}}{\frac{3}{2}+\sqrt{\frac{13}{4}}}\right)\\ & = \ln\left(\frac{18.055}{3.303}\right)\\ & = 16.986 \approx 17.0\end{align*}$$
  ::因此, 6139x2-4dx=33[osh-1(32x)]61, 并替换自然对数形式产量 6139x2-4dx=ln( 326(32262+1)-(321(32212+1))-(321(982)-(32134)=ln(98232134)=ln(18.0553.303)=16.986170)

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to determine the distance traveled along the tractix defined by  $\begin{align*}y=10 \ \text{sech}^{-1}\left(\frac{x}{10}\right)-\sqrt{10^{2}-x^{2}}\end{align*}$  between  $\begin{align*}x=10 \ ft\end{align*}$ (the start) and $\begin{align*}x=2 \ ft\end{align*}$ (the end). T
  ::早些时候,您被要求确定 y= 10 sech- 1 (x10) 102 - x2 定义的横道在 x= 10 ft (起始时间) 和 x= 2 ft (结尾时间) 之间的距离 。

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  Do you remember from an earlier concept how to compute the length of a curve, i.e., the arc length? The arc length $\begin{align*}L\end{align*}$ of the curve of a function  $\begin{align*}g(x)\end{align*}$ is computed from

  $$\begin{align*}Arc \ Length=L=\int\limits_a^b\sqrt{1+f^\prime(x)^{2}} \ dx.\end{align*}$$
  ::您是否记得, 从早先的概念中, 如何计算曲线的长度, 即弧长度 ? 函数 g( x) 的弧长度 L 是从 Arc load=L=ba1+f( x) 2 dx 计算的 。

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  Therefore,

  $$\begin{align*}L=\int\limits_1^{10}\sqrt{1+\left(\frac{d}{dx} \left [ 10\text{sech}^{-1}\left(\frac{x}{10}\right)-\sqrt{10^{2}-x^{2}} \right ]\right)^{2}}dx\\=\int\limits_1^{10}\sqrt{1+\left(-\frac{\sqrt{10^{2}-x^{2}}}{x}\right)^{2}}dx=10\ln x |_1^{10}. \end{align*}$$
  ::因此,L=1011+(ddx[10sech-1(x10)102-x2])2dx=1011+(102-x2x)2dx=10lnx(10)101。

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  The distance traveled along the curve is $\begin{align*}L=10\cdot \ln 10=23 \ ft\end{align*}$ .
  ::沿着曲线行走的距离是L=10 ln10=23英尺。

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  Example 2
  ::例2

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  Evaluate the derivative of the function $\begin{align*}\tanh^{-1}(\cos 3x)\end{align*}$ .
  ::评估函数tanh-1(cos3x)的衍生物。

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  We know that $\begin{align*}\frac{d}{dx}\tanh^{-1}u=\frac{1}{1-u^{2}}\frac{du}{dx}\end{align*}$ , so let $\begin{align*}u=\cos 3x\end{align*}$ .
  ::我们知道 ddxtanh -1u=11 -u2dudx,所以让我们 u=cos3x。

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  Then,
  ::然后,

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  $$\begin{align*}\frac{d}{dx}\tanh^{-1}(\cos 3x) & = \frac{1}{1-\cos^{2} 3x}\cdot \frac{d}{dx}\cos 3x\\ & = -\frac{3}{\sin^{2} 3x}\cdot \sin 3x\\ & = -\frac{3}{\sin 3x}\end{align*}$$
  ::ddxtanh- 1 (cos3x) = 11 - cos23xQdxxxcos3x @ 3sin23xsin3xsin3xx%3sin3xx3sin3xxx3sin3xx

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  Example 3
  ::例3

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  Evaluate the derivative of $\begin{align*}(\sinh^{-1}2x)^{\frac{3}{2}}\end{align*}$ .
  ::评价(sinh-12x)32的衍生物。

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  $$\begin{align*}\frac{d}{dx}(\sinh^{-1}2x)^{\frac{3}{2}} & = \frac{3}{2}\sqrt{\sinh^{-1}2x}\frac{d}{dx}(\sinh^{-1} 2x)\\ & = \frac{3}{2}\sqrt{\sinh^{-1}2x}\frac{1}{\sqrt{1+4x^{2}}}\cdot 2\\ & = \frac{3 \sqrt{\sinh^{-1} 2x}}{\sqrt{1+4x^{2}}}\\ & = \frac{3 \sqrt{\ln \left(2x+\sqrt{1+4x^{2}}\right)}}{\sqrt{1+4x^{2}}}\end{align*}$$
  ::ddx (sinh - 12x) 32= 32sinh - 12xx= 32sinh - 12xx) = 32sinh - 12x1=1+4x22= 3sinh - 12x1+4x2= 3sinh - 12x1=1+4x2=31+4x2=3#1+4x2

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  Example 4
  ::例4

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  Evaluate the integral $\begin{align*}\int\frac{\cos x}{\sqrt{1+\sin^{2}x}}dx\end{align*}$ .
  ::评估集成 cosx1+sin2xdx。

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  $$\begin{align*}\int \frac{\cos x}{\sqrt{1+\sin^{2} x}}dx & = \int \frac{1}{\sqrt{1+u^{2}}}du\qquad \ldots \text{Using the }u\text{-substitution:}\\ & \qquad\qquad\qquad\qquad\qquad\qquad u=\sin x \text{, and} \ du=\cos x dx\\ & = \sinh^{-1} u+C\\ & = \sinh^{-1}(\sin x)+C\\ & = \ln\left(\sin x+\sqrt{1+\sin^{2}x}\right)+C\end{align*}$$
  ::=cosx%1+sin2xxxx=11+u2du...使用u-替代:u=sinx, du=cosxdx=sinh-1u+C=sinh-1(sinx)+C=ln(sinx)1+sin2x)+C

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  Review
  ::回顾

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  For #1-7, evaluate the derivative.
  ::对于1 -7, 评估衍生物。

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  1. $\begin{align*}\sinh^{-1} 8x\end{align*}$
     ::异h- 18x
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  2. $\begin{align*}\cosh^{-1} 7x^{2}\end{align*}$
     ::cosh- 17x2
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  3. $\begin{align*}x(\tanh^{-1} 3x)^{2}\end{align*}$
     : xtanh- 13x) 2
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  4. $\begin{align*}\cosh^{-1}(\csc x)\end{align*}$
     ::COsh- 1 (cscx)
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  5. $\begin{align*}\tanh^{-1}(\cos x)\end{align*}$
     ::tanh- 1( xx)
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  6. $\begin{align*}e^{-2x}\cosh^{-1} \ 3x\end{align*}$
     ::e-2xCOS-1 3x
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  7. $\begin{align*}\text{sech}^{-1}\left ( \sqrt{3-x} \right )\end{align*}$
     ::sech- 1 (3-x)
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  8. Show that $\begin{align*}\int \frac{Adx}{\sqrt{b^{2}x^{2}-a^{2}}}=\frac{A}{b}\cosh^{-1}\left(\frac{b}{a}x\right)+C\end{align*}$ , where $\begin{align*}A,a,\end{align*}$  and $\begin{align*}b\end{align*}$  are constants.
     ::显示 = adxb2x2-a2 = Abosh-1(ax)+C, 其中 A、a 和 b 是常数 。

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

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  9. $\begin{align*}\int \frac{4}{x\sqrt{1+9x^{2}}}dx\end{align*}$
     ::4x1+9x2dx
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  10. $\begin{align*}\int \frac{\cosh x}{\sqrt{\sinh^{2} x-9}}dx\end{align*}$
      ::ososhxsinh2x-9dx(============================)
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  11. $\begin{align*}\int \frac{2x}{4-x^{4}}dx\end{align*}$ Hint: Let use $\begin{align*}u\end{align*}$ -substitution $\begin{align*}u=x^{2}\end{align*}$ .
      ::2x4 -x4dx 提示: 使用u=x2替代 u=x2 。
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  12. $\begin{align*}\int \frac{3}{(x+1)\sqrt{2x^{2}+4x+6}}dx\end{align*}$ Hint: Complete the square of the polynomial in the radical.
      ::3 (x+1) 2x2+4x+6dx Hint: 完成基数中多面体的正方形 。
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  13. $\begin{align*}\int \frac{1}{\sqrt{16x^{2}-25}}dx\end{align*}$
      ::116x2 - 25dx
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  14. $\begin{align*}\int \frac{e^{-x}}{4-e^{-2x}}dx\end{align*}$
      ::e-x4-e-2xdx
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  15. $\begin{align*}\int \frac{\cosh x}{\sinh^{2} x+4\sinh x+2} dx\end{align*}$
      ::xx+4sinhx+2dx

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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 。