课程： 7.8 涉及反向三角函数的综合体

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包含反逆三角函数的元件

The inverse trigonometric functions can be found as solutions to many problems. For some problems an inverse trigonometric function provides the angle (in radians) associated with some particular right triangle. But, for other problems, an inverse trigonometric function is a solution to a certain type of integral, and does not represent the measure of an angle. For example, an inverse trigonometric function is a solution to the following integral: $\begin{align*}\int\frac{1}{x^2-6x+13}dx\end{align*}$ . Do you know which inverse function is the solution?
::反三角函数可以作为许多问题的解决方案。对于某些问题,反三角函数提供与某个特定的右三角形相关的角度(以弧度为单位)。但是,对于其他问题,反三角函数是某类整体函数的解决方案,并不代表角度的测量。例如,反三角函数是以下整体函数的解决方案 : 1x2- 6x+13dx。你知道哪种反函数是解决方案吗?

Integrating with Inverse Trigonometric Functions
::与反向三角函数结合

Each of the derivative formulas presented in the previous concept for the inverse trigonometric functions can be associated with an integral equation. For example,
::前一概念中为逆三角函数提出的每种衍生公式都可以与一个整体等式相联系。例如,

\begin{aligned} \frac{d}{d x} [\arcsin x] = \frac{1}{\sqrt{1 - x^{2}}} \Leftrightarrow \int d [\arcsin x] = \int \frac{1}{\sqrt{1 - x^{2}}} d x = \arcsin x + C . \end{aligned}

$\begin{align*}\frac{d}{dx}[\arcsin x]=\frac{1}{\sqrt{1-x^2}}\Leftrightarrow \int d[\arcsin x] = \int \frac{1}{\sqrt{1-x^2}} dx=\arcsin x+C .\end{align*}$
::ddx[arcsinx] =11-x2}d[arcsinx] \\\\\\\\\\\ x2dx=arcsinx+C。

Applying this procedure to the derivative of each inverse trigonometric function results in these relationships:
::对每种逆三角函数的衍生物应用这一程序,导致这些关系:

**Integrals Involving Inverse Trigonometric functions: Simple Form**
$\begin{align}\int\frac{1}{\sqrt{1-x^2}}dx=\sin^{-1} x+C\end{align}$ ::11 - x2dx=sin - 1x+C	$\begin{align}\int \frac{1}{\|x\|\sqrt{x^2-1}}dx=-\csc^{-1}x+C\end{align}$
$\begin{align}\int\frac{1}{\sqrt{1-x^2}}dx=-\cos^{-1} x+C\end{align}$ ::11 - x2dx cos - 1x+C	$\begin{align}\int \frac{1}{\|x\|\sqrt{x^2-1}}dx=\sec^{-1}x+C\end{align}$
$\begin{align}\int\frac{1}{1+x^2}dx=\tan^{-1} x+C\end{align}$ ::11+x2dx=tan -1x+C	$\begin{align}\int \frac{1}{1+x^2}dx=-\cot^{-1}x+C\end{align}$

Notice that there are 3 different integrand forms. These integrand forms can be generalized to provide a larger set of integrals that can be expressed as inverse trigonometric functions using $\begin{align*}u\end{align*}$ -substitution $\begin{align*}u=\frac{b}{a}x\end{align*}$ :
::注意有三种不同的整数形式。这些整数形式可以普遍化,以提供一系列较大的整体体,可以用u-替代 u=bax表示反三角函数:

\begin{aligned} \frac{d x}{\sqrt{1 - x^{2}}} & \Rightarrow \frac{A d x}{\sqrt{a^{2} - b^{2} x^{2}}} = \frac{A}{a} \frac{d x}{\sqrt{1 - {(\frac{b}{a} x)}^{2}}} = \frac{A}{a} \frac{a}{b} \frac{d u}{\sqrt{1 - u^{2}}} = \frac{A}{b} \cdot \frac{d u}{\sqrt{1 - u^{2}}} \\ \frac{d x}{1 + x^{2}} & \Rightarrow \frac{A}{a^{2} + b^{2} x^{2}} d x = \frac{A}{a^{2}} \frac{1}{1 + {(\frac{b}{a} x)}^{2}} d x = \frac{A}{a^{2}} \cdot \frac{a}{b} \frac{d u}{1 + u^{2}} = \frac{A}{a b} \cdot \frac{d u}{1 + u^{2}} \\ \frac{d x}{| x | \sqrt{x^{2} - 1}} & \Rightarrow \frac{A d x}{| x | \sqrt{b^{2} x^{2} - a^{2}}} = \frac{A d x}{| \frac{b}{a} x | \frac{a^{2}}{b} \sqrt{{(\frac{b}{a} x)}^{2} - 1}} = \frac{A b}{a^{2}} \frac{a}{b} \frac{d u}{| u | \sqrt{u^{2} - 1}} = \frac{A}{a} \cdot \frac{d u}{| u | \sqrt{u^{2} - 1}} \end{aligned}

$\begin{align*}\frac{dx}{\sqrt{1-x^2}} &\Rightarrow \frac{Adx}{\sqrt{a^2-b^2 x^2}}=\frac{A}{a}\frac{dx}{\sqrt{1-\left(\frac{b}{a}x\right)^2}}=\frac{A}{a}\frac{a}{b}\frac{du}{\sqrt{1-u^2}} \ \qquad \qquad =\frac{A}{b}\cdot \frac{du}{\sqrt{1-u^2}}\\ \frac{dx}{1+x^2} &\Rightarrow \frac{A}{a^2+b^2x^2}dx=\frac{A}{a^2}\frac{1}{1+\left(\frac{b}{a}x\right)^2}dx=\frac{A}{a^2}\cdot\frac{a}{b}\frac{du}{1+u^2} \ \qquad \qquad =\frac{A}{ab}\cdot \frac{du}{1+u^2}\\ \frac{dx}{|x|\sqrt{x^2-1}} &\Rightarrow \frac{Adx}{|x|\sqrt{b^2x^2-a^2}}=\frac{Adx}{\Big |\frac{b}{a}x \Big |\frac{a^2}{b}\sqrt{\left(\frac{b}{a}x\right)^2-1}}=\frac{Ab}{a^2}\frac{a}{b}\frac{du}{|u|\sqrt{u^2-1}} = \frac{A}{a}\cdot \frac{du}{|u|\sqrt{u^2-1}}\end{align*}$
::dx1-x2-Adx2-B2x2=Aadx1-(ax)2=Aaabdu1-(ax)2=Aabdu1-2-u2dx1+x2*Aa2+b2x2xx=Aa2+b2x2x2x2+(ax)2+(ax)2xAa2+Aa2+(ax)2x1+Aa2x2=Aabab1+b2x2x2=Aa2+b2x2xxx2=Aa2+b2x2xxx2=Aa2+b2x2xxxx2=Aa2+(ax)2x2+(ax)2+(ax)2x2+Aa2+(ax)2+(ax)2x2+Aa2+(ax)2+Aa2+Aa2=Aa2+Aa2=Aa2*_u2=Aa_u2_1=Aa_u2_1=Aaa_

where $\begin{align*}A,a,\end{align*}$ and $\begin{align*}b\end{align*}$ are constants.
::其中A、a和b为常数。

The integrals of this generalized set of integrands can also be expressed in terms of the inverse trigonometric functions as presented in the table below:
::也可以用下表所示的逆三角函数来表示这组通用的曾数的组合:

Integrals Involving Inverse Trigonometric functions: General Form $\begin{align*}A, a,\end{align*}$ and $\begin{align*}b\end{align*}$ are constants
*$\begin{align}\int\frac{A}{\sqrt{a^2-b^2 x^2}}dx =\frac{A}{b}\sin^{-1} \frac{bx}{a}+C\end{align}$* ::Aa2-b2x2dx=Absin-1bxa+C	$\begin{align}\int\frac{A}{\|bx\|\sqrt{b^2x^2-a^2}}dx=-\frac{A}{ab}\csc^{-1}\frac{b}{a}x+C\end{align}$
*$\begin{align}\int\frac{A}{\sqrt{a^2-b^2 x^2}}dx =-\frac{A}{b}\cos^{-1} \frac{bx}{a}+C\end{align}$* ::Aa2-b2x2dxAbcos-1bxa+C	$\begin{align}\int\frac{A}{\|bx\|\sqrt{b^2x^2-a^2}}dx=\frac{A}{ab}\sec^{-1}\frac{b}{a}x+C\end{align}$
*$\begin{align}\int\frac{A}{a^2+b^2 x^2}dx= \frac{A}{ab}\tan^{-1} \frac{b}{a}x+C\end{align}$* ::a2+b2x2dx=Aabtan-1bax+C	$\begin{align}\int\frac{A}{a^2+b^2x^2}dx=-\frac{A}{ab}\cot^{-1}\frac{b}{a}x+C\end{align}$

Again, notice that $\begin{align*}u\end{align*}$ -substitution with $\begin{align*}u=\frac{b}{a}x\end{align*}$ , allows each of the general integrands to be written in the simple integrand form.
::请注意,u-替代u=bax, 允许每个普通数位数以简单数位数形式写成。

Use the information from above to evaluate the integrals:
::使用上述信息来评价组合件:

$\begin{align*}\int \frac{dx}{1+4x^2}\end{align*}$ .
::dx1+4x2 dx1+4x2 dx1+4x2.

$\begin{align*}\int \frac{5}{3x\sqrt{4x^4-1}}dx\end{align*}$ .
::-53x4x4 -1dxx。 -5x4x4x4 -1dxx。

Before we integrate $\begin{align*}\int \frac{dx}{1+4x^2}\end{align*}$ , we use $\begin{align*}u\end{align*}$ -substitution. Let $\begin{align*}u=\frac{b}{a}x=2x\end{align*}$ (the square root of $\begin{align*}4x^2\end{align*}$ ).
::在我们整合 dx1+4x2 之前, 我们使用 u 替代。 Let u =bax=2x (4x2 的平方根) 。

Then $\begin{align*}du= 2dx\end{align*}$
::然后杜=2dx

Substituting,
::以物配主者,

\begin{aligned} \int \frac{d x}{1 + 4 x^{2}} & = \int \frac{1}{1 + u^{2}} \frac{d u}{2} \\ = \frac{1}{2} \int \frac{1}{1 + u^{2}} d u \\ = \frac{1}{2} [\tan^{- 1} u + C] \\ \int \frac{d x}{1 + 4 x^{2}} & = \frac{1}{2} \tan^{- 1} 2 x + C \dots or = - \frac{1}{2} \cot^{- 1} 2 x + C \end{aligned}

$\begin{align*}\int \frac{dx}{1+4x^2} &=\int \frac{1}{1+u^2}\frac{du}{2}\\ &=\frac{1}{2}\int \frac{1}{1+u^2}du\\ &=\frac{1}{2}[\tan ^{-1}u+C]\\ \int \frac{dx}{1+4x^2} &= \frac{1}{2} \tan^{-1}2x+C \ldots \text{or} =-\frac{1}{2}\cot^{-1}2x+C\end{align*}$
::dx1+4x211+u2du2=1211+u2du=12[tan- 1u+C]dx1+4x2=12tan-12x2=12tan_12x+C...or12cot_12x+C

Before we integrate $\begin{align*}\int\frac{5}{3x\sqrt{4x^4-1}}dx\end{align*}$ , we use $\begin{align*}u\end{align*}$ -substitution. Let $\begin{align*}u=2x^2\end{align*}$ .
::在我们整合 53x4x4-1-dx之前, 我们使用 u 替代。 Let u=2x2

Then $\begin{align*}du=4xdx\end{align*}$ , or $\begin{align*}\frac{du}{4}=xdx\end{align*}$ .
::然后是4xxx或4xxx。

Substituting,
::以物配主者,

\begin{aligned} \int \frac{5 d x}{3 x \sqrt{4 x^{4} - 1}} & = \int \frac{5 x d x}{3 x^{2} \sqrt{4 x^{4} - 1}} \dots Notice the extra x in the numerator and denominator \\ to accommodate \frac{d u}{2} = x d x in the next line. \\ = \frac{5}{3} \int \frac{1}{| \frac{u}{2} | \sqrt{u^{2} - 1}} \frac{d u}{4} \\ = \frac{5}{6} \int \frac{1}{| u | \sqrt{u^{2} - 1}} d u \\ = \frac{5}{6} \sec^{- 1} u + C \\ = \frac{5}{6} \sec^{- 1} 2 x^{2} + C \end{aligned}

$\begin{align*}\int\frac{5dx}{3x\sqrt{4x^4-1}} &=\int\frac{5xdx}{3x^2\sqrt{4x^4-1}}\quad \ldots \text{Notice the extra }x \text{ in the numerator and denominator} \\ &\qquad \qquad \qquad \qquad \qquad \ \ \text{to accommodate } \frac{du}{2}=xdx \text{ in the next line.}\\ &=\frac{5}{3}\int\frac{1}{\Big | \frac{u}{2} \Big |\sqrt{u^2-1}}\frac{du}{4}\\ &=\frac{5}{6}\int\frac{1}{|u|\sqrt{u^2-1}}du\\ &=\frac{5}{6}\sec^{-1}u+C\\ &=\frac{5}{6}\sec^{-1}2x^2+C\end{align*}$
::5dx3x4x4- 15xx3x24x4- 1... 注意分子和分母中的额外 x, 以容纳 du2 =xdx 在下一行。 = 531u2u2- 1du4=561uu2- 1du=56sec- 1u+C=56sec- 12x2+C

Therefore, $\begin{align*}\int\frac{5dx}{3x\sqrt{4x^4-1}}=\frac{5}{6}\sec^{-1}2x^2+C\end{align*}$ .
::因此,5dx3x4x4-1=56sec-12x2+C。

Completing the Square
::完成广场

In some cases, an integrand that is a quadratic but is not in one of the three general forms listed above can be modified to put it in the appropriate form using the “completing the square” technique. Always look out for this possibility.
::在某些情况下,可以使用“完成广场”技术修改上文列出的三种通用形式中的一种形式,以适当形式表述为“完成广场”技术。始终注意这种可能性。

Take the integral $\begin{align*}\int\frac{5}{2x^2+4x+9}dx\end{align*}$ .
::使用积分 #%52x2+4x+9dx。

If the denominator of the integrand can be made to look like the form $\begin{align*}a^2+b^2x^2\end{align*}$ , then an inverse tangent or cotangent function can be a solution. To do so, try changing the denominator by completing the square as follows:
::如果整数位数的分母可以像表A2+b2x2一样,那么反正切或余切函数可以是一个解决办法。要做到这一点,请尝试将分母的分母修改如下:

\begin{aligned} \int \frac{5}{2 x^{2} + 4 x + 9} d x & = \frac{5}{2} \int \frac{1}{x^{2} + 2 x + \frac{9}{2}} d x \\ = \frac{5}{2} \int \frac{1}{(x^{2} + 2 x + 1) - 1 + \frac{9}{2}} d x \\ = \frac{5}{2} \int \frac{1}{(x + 1)^{2} + \frac{7}{2}} d x \dots The square has been completed. \\ Now let u = x + 1, and d u = d x . \\ = \frac{5}{2} \int \frac{1}{u^{2} + {(\sqrt{\frac{7}{2}})}^{2}} d x \\ = \frac{\frac{5}{2}}{\sqrt{\frac{7}{2}}} \tan^{- 1} \frac{1}{\sqrt{\frac{7}{2}}} u + C \\ = \frac{5 \sqrt{14}}{14} \tan^{- 1} [\frac{\sqrt{14}}{7} (x + 1)] + C \end{aligned}

$\begin{align*}\int\frac{5}{2x^2+4x+9}dx &=\frac{5}{2}\int\frac{1}{x^2+2x+\frac{9}{2}}dx\\ &=\frac{5}{2}\int\frac{1}{(x^2+2x+1)-1+\frac{9}{2}}dx\\ &=\frac{5}{2}\int\frac{1}{(x+1)^2+\frac{7}{2}}dx \qquad \qquad \ldots \text{The square has been completed.}\\ &\qquad\qquad \qquad \qquad \qquad \qquad \qquad \quad \ \text{Now let }u=x+1\text{, and } du=dx.\\ &=\frac{5}{2}\int\frac{1}{u^2+\left(\sqrt{\frac{7}{2}}\right)^2}dx\\ &=\frac{\frac{5}{2}}{\sqrt{\frac{7}{2}}}\tan^{-1}\frac{1}{\sqrt{\frac{7}{2}}}u+C\\ &=\frac{5\sqrt{14}}{14}\tan^{-1}\left[\frac{\sqrt{14}}{7}(x+1)\right]+C\end{align*}$
::52x2+4x+9dx=521x2+2x2x+92dx=521(x2+2x+1)-1+92dx=521(x+1)2+72dx...方块已完成。现在让 u=x+1, du=dx.=521u2+(722dx=5272tan-1)__1722u+C=514tan-1}[147(x+1)]+C

Examples
::实例

Example 1
::例1

Earlier, you were asked which inverse function is the solution to the integral $\begin{align*}\int\frac{1}{x^2-6x+13}dx\end{align*}$ .
::早些时候,有人问您哪个反函数是 1x2 - 6x+13dx 的组成部分的解决方案。

While the integrand may look totally unrelated to any trigonometric function or an inverse, it can be transformed so that the integral is. By employing “completing the square” and $\begin{align*}u\end{align*}$ -substitution, the integral is evaluated as follows:
::虽然百万位元看起来可能与任何三角函数或相反的函数完全无关,但可以转换成整体体。通过使用“完成广场”和“u-替代 ” 来评估整体体,其评估方式如下:

\begin{aligned} \int \frac{1}{x^{2} - 6 x + 13} d x = \int \frac{1}{(x - 3)^{2} + 4} d x = \frac{1}{4} \int \frac{1}{{(\frac{x - 3}{2})}^{2} + 1} d x = \frac{1}{2} \tan^{- 1} (\frac{x - 3}{2}) + C . \end{aligned}

$\begin{align*}\int\frac{1}{x^2-6x+13}dx=\int\frac{1}{(x-3)^2+4}dx=\frac{1}{4}\int\frac{1}{\left(\frac{x-3}{2}\right)^2+1}dx=\frac{1}{2}\tan^{-1}\left(\frac{x-3}{2}\right)+C.\end{align*}$
::1x2-6x+13dx1(x-33)2+4dx=141(x-32)2+1dx=12tan-1(x-32)+C。

Therefore, the integral solution is related to the inverse tangent.
::因此,整体解决办法与相反的偏差有关。

Example 2
::例2

Evaluate $\begin{align*}\int\frac{e^x}{\sqrt{1-e^{2x}}}dx\end{align*}$ .
::评估 ex1 - e2xdx。

Before we integrate, we use $\begin{align*}u\end{align*}$ -substitution. Let $\begin{align*}u=e^x\end{align*}$ .
::在我们融合之前,我们使用替代工具,让u=ex

Then $\begin{align*}du=e^x dx\end{align*}$ .
::然后是德克萨斯州

Substituting,
::以物配主者,

\begin{aligned} \int \frac{e^{x}}{\sqrt{1 - e^{2 x}}} d x & = \int \frac{1}{\sqrt{1 - u^{2}}} d u \\ = \sin^{- 1} u + C \dots or - \cos^{- 1} u + C \\ \int \frac{e^{x}}{\sqrt{1 - e^{2 x}}} d x & = \sin^{- 1} e^{x} + C \dots or - \cos^{- 1} e^{x} + C \end{aligned}

$\begin{align*}\int\frac{e^x}{\sqrt{1-e^{2x}}}dx & = \int\frac{1}{\sqrt{1-u^2}}du\\ &=\sin^{-1}u+C \ \ldots \text{or}-\cos^{-1}u+C\\ \int\frac{e^x}{\sqrt{1-e^{2x}}}dx &=\sin^{-1}e^x+C \ldots \text{or}-\cos^{-1}e^x+C\\\end{align*}$
::ex1 - e2xxx11 - u2du=sin - 1u+C... 或 - cos - 1u+C ex1 - e2xx=sin - 1ex+C... or - cos - 1ex+C

Example 3
::例3

Evaluate the definite integral $\begin{align*}\int\limits_{\ln 2}^{\ln \left(\frac{2}{\sqrt{3}}\right)}\frac{e^{-x}}{\sqrt{1-e^{-2x}}}dx\end{align*}$ .
::评估确定的整体(23)e-x1-e-2xdx。

With $\begin{align*}u\end{align*}$ -substitution, $\begin{align*}u=e^{-x}\end{align*}$ and $\begin{align*}du=-e^{-x}dx\end{align*}$ .
::与u-替代,u=e-x 和 du-e-xdx。

The limits of integration are now: $\begin{align*}u=e^{-\ln 2} =\frac{1}{2}\end{align*}$ and $\begin{align*}u=e^{-\ln\frac{2}{\sqrt{3}}}=\frac{\sqrt{3}}{2}\end{align*}$ .
::整合的极限是:u=e-ln2=12和u=e-ln23=32。

Thus our integral becomes
::如此一来,我们整体就变成

\begin{aligned} \int_{\ln 2}^{\ln (\frac{2}{\sqrt{3}})} \frac{e^{- x}}{\sqrt{1 - e^{- 2 x}}} d x & = \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \frac{- u}{\sqrt{1 - u^{2}}} \frac{d u}{u} \\ = \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \frac{1}{\sqrt{1 - u^{2}}} d u \\ = - \sin^{- 1} u |_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \\ = - \sin^{- 1} (\frac{\sqrt{3}}{2}) + \sin^{- 1} (\frac{1}{2}) \\ = - \frac{π}{3} + \frac{π}{6} \\ = - \frac{π}{6} \end{aligned}

$\begin{align*}\int_{\ln 2}^{\ln\left(\frac{2}{\sqrt{3}}\right)}\frac{e^{-x}}{\sqrt{1-e^{-2x}}}dx &=\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{-u}{\sqrt{1-u^2}}\frac{du}{u}\\ &=\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{1}{\sqrt{1-u^2}}du\\ &=-\sin^{-1}u \Big |_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\\ &=-\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)+\sin^{-1}\left(\frac{1}{2}\right)\\ &=-\frac{\pi}{3}+\frac{\pi}{6}\\ &=-\frac{\pi}{6}\end{align*}$
::=====================================================================================================================================================

Therefore, $\begin{align*}\int\limits_{\ln 2}^{\ln\left(\frac{2}{\sqrt{3}}\right)}\frac{e^{-x}}{\sqrt{1-e^{-2x}}}dx=-\frac{\pi}{6}\end{align*}$ .
::因此,(23)-x1-e-2xdx_#6。

Review
::回顾

For #1-4, evaluate the integral.
::14号,评估整体体

$\begin{align*}\int\frac{dx}{\sqrt{9-x^2}}\end{align*}$
::* dx9 - x2

$\begin{align*}\int\limits_{1}^{3}\frac{dx}{\sqrt{x}(x+1)}\end{align*}$
::13dxxx(x+1)

$\begin{align*}\int\frac{x-3}{x^2+1}dx\end{align*}$
::X - 3x2+1dx

$\begin{align*}\int\limits_{-\sqrt{3}}^{0}\frac{x}{1+x^2}dx\end{align*}$
::30x1+x2dx

Given the points $\begin{align*}A(2,1)\end{align*}$ and $\begin{align*}B(5,4)\end{align*}$ , find a point $\begin{align*}Q\end{align*}$ in the interval $\begin{align*}[2,5]\end{align*}$ on the $\begin{align*}x\end{align*}$ -axis that maximizes angle $\begin{align*}\measuredangle AQB\end{align*}$ .
::鉴于 A(2,1) 和 B(5,4) 点,在 X 轴的间隔[2,5] 中找到一个Q点,使角最大化 AQB。

Evaluate the integral $\begin{align*}\int\frac{2}{x\sqrt{x^2-1}}dx\end{align*}$
::评估集成 2x22-1dx

Evaluate the integral $\begin{align*}\int\frac{5x+4}{9x^2+1}dx\end{align*}$
::评估 #% 5x+49x2+1dx

Find the volume of the solid generated by rotating the region bounded by $\begin{align*}y=\frac{1}{\sqrt{1+4x^2}}\end{align*}$ , $\begin{align*}x=0\end{align*}$ , and $\begin{align*}x=\frac{\sqrt{3}}{2}\end{align*}$ about the $\begin{align*}x\end{align*}$ -axis.
::查找Y=11+4x2、x=0和x=32的 X轴区域旋转产生的固体体积。

Evaluate the integral $\begin{align*}\int\frac{e^{2x}}{7+e^{4x}}dx\end{align*}$ , Hint: Use $\begin{align*}u=e^{2x}\end{align*}$ .
::评估 {e2x7+e4xxx, 提示: 使用 u=e2x。

Evaluate the integral $\begin{align*}\int\frac{3}{x^2+8x+17}dx\end{align*}$ , Hint: First complete the square for the polynomial in the denominator; then use $\begin{align*}u\end{align*}$ -substitution for the square in the denominator.
::评估 {% 3x2+8x+17dx, 提示 : 首先完成分母中多数值方块的方块; 然后对分母中的方块使用 u 替代。

Find the area bounded by the curve $\begin{align*}\sqrt{9-x^2}\cdot y=1\end{align*}$ and the lines $\begin{align*}x=0\end{align*}$ , $\begin{align*}x=2\end{align*}$ and $\begin{align*}y=0\end{align*}$ .
::查找曲线 9 - x2y=1 和线条 x=0, x=2 和 y=0 之间的区域。

Evaluate the integral $\begin{align*}\int\frac{x+2}{\sqrt{4-x^2}}dx\end{align*}$ ; Hint: Breakup the integrand.
::评估 {x+24_x2dx; 提示: 分手整数。

Evaluate the integral $\begin{align*}\int\frac{3x^5}{x^4+1}dx\end{align*}$
::评估 {% 3x5x4+1dx

Evaluate the integral $\begin{align*}\int \frac{5e^x}{\sqrt{81-4e^{2x}}}dx\end{align*}$
::评估 5ex81 - 4e2xxx

Evaluate the integral $\begin{align*}\int\frac{2x+5}{x^2-4x+6}dx\end{align*}$ Hint: Break up the numerator into two parts whose sum equals the derivative of the denominator.
::评估 #% 2x+5x2-4x+6dx 提示:将分子分成两部分,两部分的总和等于分母的衍生物。

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$\begin{align}\int\frac{1}{\sqrt{1-x^2}}dx=\sin^{-1} x+C\end{align}$ ::11 - x2dx=sin - 1x+C	$\begin{align}\int \frac{1}{\|x\|\sqrt{x^2-1}}dx=-\csc^{-1}x+C\end{align}$
$\begin{align}\int\frac{1}{\sqrt{1-x^2}}dx=-\cos^{-1} x+C\end{align}$ ::11 - x2dx cos - 1x+C	$\begin{align}\int \frac{1}{\|x\|\sqrt{x^2-1}}dx=\sec^{-1}x+C\end{align}$
$\begin{align}\int\frac{1}{1+x^2}dx=\tan^{-1} x+C\end{align}$ ::11+x2dx=tan -1x+C	$\begin{align}\int \frac{1}{1+x^2}dx=-\cot^{-1}x+C\end{align}$

*$\begin{align}\int\frac{A}{\sqrt{a^2-b^2 x^2}}dx =\frac{A}{b}\sin^{-1} \frac{bx}{a}+C\end{align}$* ::Aa2-b2x2dx=Absin-1bxa+C	$\begin{align}\int\frac{A}{\|bx\|\sqrt{b^2x^2-a^2}}dx=-\frac{A}{ab}\csc^{-1}\frac{b}{a}x+C\end{align}$
*$\begin{align}\int\frac{A}{\sqrt{a^2-b^2 x^2}}dx =-\frac{A}{b}\cos^{-1} \frac{bx}{a}+C\end{align}$* ::Aa2-b2x2dxAbcos-1bxa+C	$\begin{align}\int\frac{A}{\|bx\|\sqrt{b^2x^2-a^2}}dx=\frac{A}{ab}\sec^{-1}\frac{b}{a}x+C\end{align}$
*$\begin{align}\int\frac{A}{a^2+b^2 x^2}dx= \frac{A}{ab}\tan^{-1} \frac{b}{a}x+C\end{align}$* ::a2+b2x2dx=Aabtan-1bax+C	$\begin{align}\int\frac{A}{a^2+b^2x^2}dx=-\frac{A}{ab}\cot^{-1}\frac{b}{a}x+C\end{align}$

章节大纲

常规

包含反逆三角函数的元件

Integrating with Inverse Trigonometric Functions ::与反向三角函数结合

Completing the Square ::完成广场

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

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Review (Answers) ::回顾(答复)

Integrating with Inverse Trigonometric Functions
::与反向三角函数结合

Completing the Square
::完成广场

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Review
::回顾

Review (Answers)
::回顾(答复)