如果你也在 怎样代写微积分Calculus 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。微积分Calculus 基本上就是非常高级的代数和几何。从某种意义上说，它甚至不是一门新学科——它采用代数和几何的普通规则，并对它们进行调整，以便它们可以用于更复杂的问题。(当然，问题在于，从另一种意义上说，这是一门新的、更困难的学科。)
微积分Calculus数学之所以有效，是因为曲线在局部是直的;换句话说，它们在微观层面上是直的。地球是圆的，但对我们来说，它看起来是平的，因为与地球的大小相比，我们在微观层面上。微积分之所以有用，是因为当你放大曲线，曲线变直时，你可以用正则代数和几何来处理它们。这种放大过程是通过极限数学来实现的。
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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
三角函数替换是指用三角函数替换积分变量。最常见的替换是$x=a \tan \theta, x=a \sin \theta$和$x=a \sec \theta$。这些替换在将涉及$\sqrt{a^2+x^2}, \sqrt{a^2-x^2}$和$\sqrt{x^2-a^2}$的积分转换为我们可以直接计算的积分时是有效的，因为它们来自图8.2中的参考直角三角形。
通过$x=a \tan \theta$，
$$
a^2+x^2=a^2+a^2 \tan ^2 \theta=a^2\left(1+\tan ^2 \theta\right)=a^2 \sec ^2 \theta .
$$
通过$x=a \sin \theta$，
$$
a^2-x^2=a^2-a^2 \sin ^2 \theta=a^2\left(1-\sin ^2 \theta\right)=a^2 \cos ^2 \theta .
$$
通过$x=a \sec \theta$，
$$
x^2-a^2=a^2 \sec ^2 \theta-a^2=a^2\left(\sec ^2 \theta-1\right)=a^2 \tan ^2 \theta .
$$
我们希望我们在积分中使用的任何替换都是可逆的，这样我们就可以在之后把它换回原来的变量。例如，如果是$x=a \tan \theta$，我们希望能够在集成发生后设置$\theta=\tan ^{-1}(x / a)$。对于$x=a \sin \theta$，我们希望在完成后能够设置$\theta=\sin ^{-1}(x / a)$，对于$x=a \sec \theta$也是如此。
正如我们在7.6节中所知道的，这些替换中的函数只对$\theta$的选定值有逆(图8.3)。对于可逆性，
$$
\begin{aligned}
& x=a \tan \theta \text { requires } \theta=\tan ^{-1}\left(\frac{x}{a}\right) \text { with }-\frac{\pi}{2}<\theta<\frac{\pi}{2}, \
& x=a \sin \theta \text { requires } \theta=\sin ^{-1}\left(\frac{x}{a}\right) \text { with }-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}, \
& x=a \sec \theta \text { requires } \theta=\sec ^{-1}\left(\frac{x}{a}\right) \text { with }\left{\begin{array}{l}
0 \leq \theta<\frac{\pi}{2} \text { if } \frac{x}{a} \geq 1, \
\frac{\pi}{2}<\theta \leq \pi \text { if } \frac{x}{a} \leq-1 .
\end{array}\right.
\end{aligned}
$$
数学代写|微积分代写Calculus代写|Integration of Rational Functions by Partial Fractions
本节展示如何将有理函数(多项式的商)表示为易于积分的简单分数(称为部分分数)的和。例如，有理函数$(5 x-3) /\left(x^2-2 x-3\right)$可以重写为
$$
\frac{5 x-3}{x^2-2 x-3}=\frac{2}{x+1}+\frac{3}{x-3} \text {. }
$$
你可以用代数方法验证这个等式，把分数放在右边的公分母$(x+1)(x-3)$上。将有理函数写成这样的和的技巧在其他情况下也很有用(例如，当使用某些变换方法来解微分方程时)。要对前面表达式左边的有理函数$(5 x-3) /\left(x^2-2 x-3\right)$积分，我们只需对右边分数的积分求和:
$$
\begin{aligned}
\int \frac{5 x-3}{(x+1)(x-3)} d x & =\int \frac{2}{x+1} d x+\int \frac{3}{x-3} d x \
& =2 \ln |x+1|+3 \ln |x-3|+C .
\end{aligned}
$$
把有理函数写成简单分数和的方法叫做部分分数法。在前面的例子中，它包括查找常量$A$和$B$，以便
$$
\frac{5 x-3}{x^2-2 x-3}=\frac{A}{x+1}+\frac{B}{x-3} .
$$
(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。
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