## 数学代写|微积分代写Calculus代写|MTH-211

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## 数学代写|微积分代写Calculus代写|Trigonometric Substitutions

Trigonometric substitutions occur when we replace the variable of integration by a trigonometric function. The most common substitutions are $x=a \tan \theta, x=a \sin \theta$, and $x=a \sec \theta$. These substitutions are effective in transforming integrals involving $\sqrt{a^2+x^2}, \sqrt{a^2-x^2}$, and $\sqrt{x^2-a^2}$ into integrals we can evaluate directly since they come from the reference right triangles in Figure 8.2.

With $x=a \tan \theta$,
$$a^2+x^2=a^2+a^2 \tan ^2 \theta=a^2\left(1+\tan ^2 \theta\right)=a^2 \sec ^2 \theta .$$
With $x=a \sin \theta$,
$$a^2-x^2=a^2-a^2 \sin ^2 \theta=a^2\left(1-\sin ^2 \theta\right)=a^2 \cos ^2 \theta .$$
With $x=a \sec \theta$,
$$x^2-a^2=a^2 \sec ^2 \theta-a^2=a^2\left(\sec ^2 \theta-1\right)=a^2 \tan ^2 \theta .$$
We want any substitution we use in an integration to be reversible so that we can change back to the original variable afterward. For example, if $x=a \tan \theta$, we want to be able to set $\theta=\tan ^{-1}(x / a)$ after the integration takes place. If $x=a \sin \theta$, we want to be able to set $\theta=\sin ^{-1}(x / a)$ when we’re done, and similarly for $x=a \sec \theta$.

As we know from Section 7.6, the functions in these substitutions have inverses only for selected values of $\theta$ (Figure 8.3). For reversibility,
\begin{aligned} & x=a \tan \theta \text { requires } \theta=\tan ^{-1}\left(\frac{x}{a}\right) \text { with }-\frac{\pi}{2}<\theta<\frac{\pi}{2}, \ & x=a \sin \theta \text { requires } \theta=\sin ^{-1}\left(\frac{x}{a}\right) \text { with }-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}, \ & x=a \sec \theta \text { requires } \theta=\sec ^{-1}\left(\frac{x}{a}\right) \text { with }\left{\begin{array}{l} 0 \leq \theta<\frac{\pi}{2} \text { if } \frac{x}{a} \geq 1, \ \frac{\pi}{2}<\theta \leq \pi \text { if } \frac{x}{a} \leq-1 . \end{array}\right. \end{aligned}

## 数学代写|微积分代写Calculus代写|Integration of Rational Functions by Partial Fractions

This section shows how to express a rational function (a quotient of polynomials) as a sum of simpler fractions, called partial fractions, which are easily integrated. For instance, the rational function $(5 x-3) /\left(x^2-2 x-3\right)$ can be rewritten as
$$\frac{5 x-3}{x^2-2 x-3}=\frac{2}{x+1}+\frac{3}{x-3} \text {. }$$
You can verify this equation algebraically by placing the fractions on the right side over a common denominator $(x+1)(x-3)$. The skill acquired in writing rational functions as such a sum is useful in other settings as well (for instance, when using certain transform methods to solve differential equations). To integrate the rational function $(5 x-3) /\left(x^2-2 x-3\right)$ on the left side of our previous expression, we simply sum the integrals of the fractions on the right side:
\begin{aligned} \int \frac{5 x-3}{(x+1)(x-3)} d x & =\int \frac{2}{x+1} d x+\int \frac{3}{x-3} d x \ & =2 \ln |x+1|+3 \ln |x-3|+C . \end{aligned}
The method for rewriting rational functions as a sum of simpler fractions is called the method of partial fractions. In the case of the preceding example, it consists of finding constants $A$ and $B$ such that
$$\frac{5 x-3}{x^2-2 x-3}=\frac{A}{x+1}+\frac{B}{x-3} .$$
(Pretend for a moment that we do not know that $A=2$ and $B=3$ will work.) We call the fractions $A /(x+1)$ and $B /(x-3)$ partial fractions because their denominators are only part of the original denominator $x^2-2 x-3$. We call $A$ and $B$ undetermined coefficients until suitable values for them have been found.

# 微积分代考

## 数学代写|微积分代写Calculus代写|Trigonometric Substitutions

$$a^2+x^2=a^2+a^2 \tan ^2 \theta=a^2\left(1+\tan ^2 \theta\right)=a^2 \sec ^2 \theta .$$

$$a^2-x^2=a^2-a^2 \sin ^2 \theta=a^2\left(1-\sin ^2 \theta\right)=a^2 \cos ^2 \theta .$$

$$x^2-a^2=a^2 \sec ^2 \theta-a^2=a^2\left(\sec ^2 \theta-1\right)=a^2 \tan ^2 \theta .$$

\begin{aligned} & x=a \tan \theta \text { requires } \theta=\tan ^{-1}\left(\frac{x}{a}\right) \text { with }-\frac{\pi}{2}<\theta<\frac{\pi}{2}, \ & x=a \sin \theta \text { requires } \theta=\sin ^{-1}\left(\frac{x}{a}\right) \text { with }-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}, \ & x=a \sec \theta \text { requires } \theta=\sec ^{-1}\left(\frac{x}{a}\right) \text { with }\left{\begin{array}{l} 0 \leq \theta<\frac{\pi}{2} \text { if } \frac{x}{a} \geq 1, \ \frac{\pi}{2}<\theta \leq \pi \text { if } \frac{x}{a} \leq-1 . \end{array}\right. \end{aligned}

## 数学代写|微积分代写Calculus代写|Integration of Rational Functions by Partial Fractions

$$\frac{5 x-3}{x^2-2 x-3}=\frac{2}{x+1}+\frac{3}{x-3} \text {. }$$

\begin{aligned} \int \frac{5 x-3}{(x+1)(x-3)} d x & =\int \frac{2}{x+1} d x+\int \frac{3}{x-3} d x \ & =2 \ln |x+1|+3 \ln |x-3|+C . \end{aligned}

$$\frac{5 x-3}{x^2-2 x-3}=\frac{A}{x+1}+\frac{B}{x-3} .$$
(暂时假设我们不知道$A=2$和$B=3$会正常工作。)我们称分数为$A /(x+1)$和$B /(x-3)$部分分数因为它们的分母只是原分母$x^2-2 x-3$的一部分。我们称$A$和$B$为待定系数，直到找到适合它们的值。

## 数学代写|微积分代写Calculus代写|MTH251

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## 数学代写|微积分代写Calculus代写|Inverse Hyperbolic Functions

The inverses of the six basic hyperbolic functions are very useful in integration . Since $d(\sinh x) / d x=\cosh x>0$, the hyperbolic sine is an increasing function of $x$. We denote its inverse by
$$y=\sinh ^{-1} x .$$
For every value of $x$ in the interval $-\infty<x<\infty$, the value of $y=\sinh ^{-1} x$ is the number whose hyperbolic sine is $x$. The graphs of $y=\sinh x$ and $y=\sinh ^{-1} x$ are shown in Figure 7.32a.

The function $y=\cosh x$ is not one-to-one because its graph in Table 7.5 does not pass the horizontal line test. The restricted function $y=\cosh x, x \geq 0$, however, is oneto-one and therefore has an inverse, denoted by
$$y=\cosh ^{-1} x .$$

For every value of $x \geq 1, y=\cosh ^{-1} x$ is the number in the interval $0 \leq y<\infty$ whose hyperbolic cosine is $x$. The graphs of $y=\cosh x, x \geq 0$, and $y=\cosh ^{-1} x$ are shown in Figure 7.32b.

Like $y=\cosh x$, the function $y=\operatorname{sech} x=1 / \cosh x$ fails to be one-to-one, but its restriction to nonnegative values of $x$ does have an inverse, denoted by
$$y=\operatorname{sech}^{-1} x .$$
For every value of $x$ in the interval $(0,1], y=\operatorname{sech}^{-1} x$ is the nonnegative number whose hyperbolic secant is $x$. The graphs of $y=\operatorname{sech} x, x \geq 0$, and $y=\operatorname{sech}^{-1} x$ are shown in Figure 7.32c.

The hyperbolic tangent, cotangent, and cosecant are one-to-one on their domains and therefore have inverses, denoted by
$$y=\tanh ^{-1} x, \quad y=\operatorname{coth}^{-1} x, \quad y=\operatorname{csch}^{-1} x .$$
These functions are graphed in Figure 7.33.

## 数学代写|微积分代写Calculus代写|Useful Identities

We use the identities in Table 7.9 to calculate the values of $\operatorname{sech}^{-1} x, \operatorname{csch}^{-1} x$, and $\operatorname{coth}^{-1} x$ on calculators that give only $\cosh ^{-1} x, \sinh ^{-1} x$, and $\tanh ^{-1} x$. These identities are direct consequences of the definitions. For example, if $0<x \leq 1$, then
$$\operatorname{sech}\left(\cosh ^{-1}\left(\frac{1}{x}\right)\right)=\frac{1}{\cosh \left(\cosh ^{-1}\left(\frac{1}{x}\right)\right)}=\frac{1}{\left(\frac{1}{x}\right)}=x .$$
We also know that $\operatorname{sech}\left(\operatorname{sech}^{-1} x\right)=x$, so because the hyperbolic secant is one-to-one on $(0,1]$, we have
$$\cosh ^{-1}\left(\frac{1}{x}\right)=\operatorname{sech}^{-1} x$$

# 微积分代考

## 数学代写|微积分代写Calculus代写|Inverse Hyperbolic Functions

$$y=\sinh ^{-1} x .$$

$$y=\cosh ^{-1} x .$$

$$y=\operatorname{sech}^{-1} x .$$

$$y=\tanh ^{-1} x, \quad y=\operatorname{coth}^{-1} x, \quad y=\operatorname{csch}^{-1} x .$$

## 数学代写|微积分代写Calculus代写|Useful Identities

$$\operatorname{sech}\left(\cosh ^{-1}\left(\frac{1}{x}\right)\right)=\frac{1}{\cosh \left(\cosh ^{-1}\left(\frac{1}{x}\right)\right)}=\frac{1}{\left(\frac{1}{x}\right)}=x .$$

$$\cosh ^{-1}\left(\frac{1}{x}\right)=\operatorname{sech}^{-1} x$$

## 数学代写|微积分代写Calculus代写|MATH2310

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## 数学代写|微积分代写Calculus代写|Tricky Trig Integrals

In this section, you integrate powers of the six trigonometric functions, like $\int \sin ^3(x) d x$ and $\int \sec ^4(x) d x$, and products or quotients of trig functions, like $\int \sin ^2(x) \cos ^3(x) d x$ and $\int \frac{\csc ^2(x)}{\cot (x)} d x$. This is a bit tedious – time for some caffeine. To use the following techniques, you must have an integrand that contains just one of the six trig functions like $\int \csc ^3(x) d x$ or a certain pairing of trig functions, like $\int \sin ^2(x) \cos (x) d x$. If the integrand has two trig functions, the two must be one of these three pairs: sine with cosine, secant with tangent, or cosecant with cotangent. For an integrand containing something other than one of these pairs, you can convert the problem into one of these pairs by using trig identities like $\sin (x)=1 / \csc (x)$ and $\tan (x)=\sin (x) / \cos (x)$. For instance,
\begin{aligned} & \int \sin ^2(x) \sec (x) \tan (x) d x \ = & \int \sin ^2(x) \frac{1}{\cos (x)} \cdot \frac{\sin (x)}{\cos (x)} d x \ = & \int \frac{\sin ^3(x)}{\cos ^2(x)} d x \end{aligned}
After any needed conversions, you get one of three cases:
\begin{aligned} & \int \sin ^m(x) \cos ^n(x) d x \ & \int \sec ^m(x) \tan ^n(x) d x \ & \int \csc ^m(x) \cot ^n(x) d x \end{aligned}
where either $m$ or $n$ is a positive integer.

## 数学代写|微积分代写Calculus代写|Sines and cosines

Case 1: The power of sine is odd and positive
If the power of sine is odd and positive, lop off one sine factor and put it to the right of the rest of the expression, convert the remaining sine factors to cosines with the Pythagorean identity, and then integrate with the substitution method where $u=\cos (x)$.
The Pythagorean identity tells you that, for any angle $x, \sin ^2(x)+$ $\cos ^2(x)=1$. And thus $\sin ^2(x)=1-\cos ^2(x)$ and $\cos ^2(x)=1-$ $\sin ^2(x)$.
Now integrate $\int \sin ^3(x) \cos ^4(x) d x$.

1. Lop off one sine factor and move it to the right.
$$\int \sin ^3(x) \cos ^4(x) d x=\int \sin ^2(x) \cos ^4(x) \sin (x) d x$$
2. Convert the remaining sines to cosines using the Pythagorean identity and simplify.
\begin{aligned} & \int \sin ^2(x) \cos ^4(x) \sin (x) d x \ = & \int\left(1-\cos ^2(x)\right) \cos ^4(x) \sin (x) d x \ = & \int\left(\cos ^4(x)-\cos ^6(x)\right) \sin (x) d x \end{aligned}
3. Integrate with substitution, where $u=\cos (x)$.
$$\begin{gathered} u=\cos (x) \ \frac{d u}{d x}=-\sin (x) \ d u=-\sin (x) d x \end{gathered}$$

# 微积分代考

## 数学代写|微积分代写Calculus代写|Tricky Trig Integrals

\begin{aligned} & \int \sin ^2(x) \sec (x) \tan (x) d x \ = & \int \sin ^2(x) \frac{1}{\cos (x)} \cdot \frac{\sin (x)}{\cos (x)} d x \ = & \int \frac{\sin ^3(x)}{\cos ^2(x)} d x \end{aligned}

\begin{aligned} & \int \sin ^m(x) \cos ^n(x) d x \ & \int \sec ^m(x) \tan ^n(x) d x \ & \int \csc ^m(x) \cot ^n(x) d x \end{aligned}

## 数学代写|微积分代写Calculus代写|Sines and cosines

$$\int \sin ^3(x) \cos ^4(x) d x=\int \sin ^2(x) \cos ^4(x) \sin (x) d x$$

\begin{aligned} & \int \sin ^2(x) \cos ^4(x) \sin (x) d x \ = & \int\left(1-\cos ^2(x)\right) \cos ^4(x) \sin (x) d x \ = & \int\left(\cos ^4(x)-\cos ^6(x)\right) \sin (x) d x \end{aligned}

$$\begin{gathered} u=\cos (x) \ \frac{d u}{d x}=-\sin (x) \ d u=-\sin (x) d x \end{gathered}$$

## 数学代写|微积分代写Calculus代写|Radioactivity

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## 数学代写|微积分代写Calculus代写|Radioactivity

Some atoms are unstable and can spontaneously emit mass or radiation. This process is called radioactive decay, and an element whose atoms go spontaneously through this process is called radioactive. Sometimes when an atom emits some of its mass through this process of radioactivity, the remainder of the atom re-forms to make an atom of some new element. For example, radioactive carbon-14 decays into nitrogen; radium, through a number of intermediate radioactive steps, decays into lead.

Experiments have shown that at any given time the rate at which a radioactive element decays (as measured by the number of nuclei that change per unit time) is approximately proportional to the number of radioactive nuclei present. Thus, the decay of a radioactive element is described by the equation $d y / d t=-k y, k>0$. It is conventional to use $-k$, with $k>0$, to emphasize that $y$ is decreasing. If $y_0$ is the number of radioactive nuclei present at time zero, the number still present at any later time $t$ will be
$$y=y_0 e^{-k t}, \quad k>0 .$$
The half-life of a radioactive element is the time expected to pass until half of the radioactive nuclei present in a sample decay. It is an interesting fact that the half-life is a constant that does not depend on the number of radioactive nuclei initially present in the sample, but only on the radioactive substance.

To compute the half-life, let $y_0$ be the number of radioactive nuclei initially present in the sample. Then the number $y$ present at any later time $t$ will be $y=y_0 e^{-k t}$. We seek the value of $t$ at which the number of radioactive nuclei present equals half the original number:
\begin{aligned} y_0 e^{-k t} & =\frac{1}{2} y_0 \ e^{-k t} & =\frac{1}{2} \ -k t & =\ln \frac{1}{2}=-\ln 2 \quad \text { Reciprocal Rule for logarithms } \ t & =\frac{\ln 2}{k} . \end{aligned}

## 数学代写|微积分代写Calculus代写|Heat Transfer: Newton’s Law of Cooling

Hot soup left in a tin cup cools to the temperature of the surrounding air. A hot silver bar immersed in a large tub of water cools to the temperature of the surrounding water. In situations like these, the rate at which an object’s temperature is changing at any given time is roughly proportional to the difference between its temperature and the temperature of the surrounding medium. This observation is called Newton’s Law of Cooling, although it applies to warming as well.

If $H$ is the temperature of the object at time $t$ and $H_S$ is the constant surrounding temperature, then the differential equation is
$$\frac{d H}{d t}=-k\left(H-H_S\right)$$
If we substitute $y$ for $\left(H-H_S\right)$, then
$$\begin{array}{rlrl} \frac{d y}{d t} & =\frac{d}{d t}\left(H-H_S\right)=\frac{d H}{d t}-\frac{d}{d t}\left(H_S\right) & \ & =\frac{d H}{d t}-0 & & \ & =\frac{d H}{d t} & & H_S \text { is a constant. } \ & =-k\left(H-H_S\right) & \ & =-k y . & & \text { Eq. (8) } \ & & H-H_S=y \end{array}$$
We know that the solution of the equation $d y / d t=-k y$ is $y=y_0 e^{-k t}$, where $y(0)=y_0$. Substituting $\left(H-H_S\right)$ for $y$, this says that
$$H-H_S=\left(H_0-H_S\right) e^{-k t},$$
where $H_0$ is the temperature at $t=0$. This equation is the solution to Newton’s Law of Cooling.

# 微积分代考

## 数学代写|微积分代写Calculus代写|Radioactivity

$$y=y_0 e^{-k t}, \quad k>0 .$$

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。