# 积分（FP3）

> 爱德思国际A-Level进阶数学 · 2018版IAL FP3（WFM03）
> 来源: https://www.owlsprep.com/zh/study/edexcel-ial-further-math-fp3-integration/

本指南覆盖爱德思IAL FP3全部6个核心积分考点，包括双曲函数/反函数积分、换元法、递推降阶公式、弧长与旋转曲面面积，附带分步详解例题与针对性考试技巧。

**先修:** [分部积分法与换元积分法（IAL A-Level数学纯数3内容）](https://www.owlsprep.com/zh/study/edexcel-ial-math-p3-integration/); [双曲函数与反函数的求导（FP3微分章节内容）](https://www.owlsprep.com/zh/study/edexcel-ial-further-math-fp3-differentiation/)

## 学习目标

- 利用标准积分结果与分部积分法对双曲函数、反三角函数与反双曲函数进行积分运算
- 运用三角/双曲换元法求解标准倒数形式与根式形式的积分
- 推导并应用定积分与不定积分的递推降阶公式
- 计算直角坐标曲线与参数曲线的弧长和旋转曲面面积

## 双曲函数、反三角函数与反双曲函数的积分

双曲函数的标准积分结果已在公式手册中给出，你可以直接应用它们求解不定积分与定积分。对于反三角函数与反双曲函数，使用分部积分法，令$u$等于该反函数，$dv = dx$。

**反函数积分** — 对于任意可积反函数$f^{-1}(x)$，分部积分的设置为：$u = f^{-1}(x)$，$dv = dx$，因此$du = \frac{1}{f'(f^{-1}(x))}dx$，$v = x$，可得$\int f^{-1}(x)dx = x f^{-1}(x) - \int \frac{x}{f'(f^{-1}(x))}dx$。

**例题:** 计算$\int_0^1 \text{arsinh } x dx$

1. 令$u = \text{arsinh }x$，$dv = dx$
2. $$du = \frac{1}{\sqrt{1+x^2}}dx, v = x$$
3. 代入分部积分公式：
4. $$\int \text{arsinh }x dx = x \text{arsinh }x - \int \frac{x}{\sqrt{1+x^2}}dx$$
5. 使用换元$t = 1+x^2$，$dt = 2x dx$计算剩余积分：
6. $$\int \frac{x}{\sqrt{1+x^2}}dx = \sqrt{1+x^2} + C$$
7. 代入上下限0到1：
8. $$[x \text{arsinh }x - \sqrt{1+x^2}]_0^1 = \text{arsinh }1 - \sqrt{2} + 1$$
9. 利用标准结果$\text{arsinh }1 = \ln(1 + \sqrt{2})$得到最终值：$\ln(1+\sqrt{2}) - \sqrt{2} + 1 \approx 0.467$

> **考试技巧**
>
> 在应用分部积分法后，你可以直接使用公式手册中的标准根式积分结果计算剩余积分，以节省考试时间。

*计算器:* allowed

## 积分的三角换元与双曲换元

使用标准换元简化包含二次根式的积分。所有标准换元与对应的积分结果都已在公式手册中给出。对于更复杂的根式积分，题目会给出所需的换元方式，你无需自行构造新的换元。

| Integrand Form | Substitution | Simplified Surd |
| --- | --- | --- |
| $1/\sqrt{a^2 + x^2}$ | $x = a \sinh u$ | $a \cosh u$ |
| $1/\sqrt{a^2 - x^2}$ | $x = a \sin \theta$ | $a \cos \theta$ |
| $1/\sqrt{x^2 - a^2}$ | $x = a \cosh u$ | $a \sinh u$ |
| $1/(a^2 + x^2)$ | $x = a \tan \theta$ | $a^2 \sec^2 \theta$ |

**例题:** 当$x > 3/2$时，计算$\int \frac{1}{\sqrt{4x^2 - 9}}dx$

1. 将被积函数改写为匹配标准形式的样式：
2. $$\int \frac{1}{\sqrt{(2x)^2 - 3^2}}dx$$
3. 使用换元$2x = 3 \cosh u$，因此$2dx = 3 \sinh u du$，$dx = \frac{3}{2}\sinh u du$
4. 将换元代入积分：
5. $$\int \frac{1}{\sqrt{9 \cosh^2 u - 9}} \times \frac{3}{2}\sinh u du = \int \frac{1}{3 \sinh u} \times \frac{3}{2}\sinh u du = \int \frac{1}{2} du$$
6. $$= \frac{1}{2}u + C = \frac{1}{2}\text{arcosh}\left(\frac{2x}{3}\right) + C$$
7. 该结果与公式手册中的标准结果一致，因此你也可以直接写出结果为$\frac{1}{2}\ln\left(2x + \sqrt{4x^2 - 9}\right) + C$

*计算器:* allowed

## 递推降阶公式的推导与应用

**考试命令词**

递推降阶公式类题目的指令词有明确的考察要求：

- **Derive** — Show full working to get the recurrence relation from the definition of $I_n$, no marks are awarded for stating the relation without proof

- **Evaluate** — Apply the derived recurrence relation to base cases (usually $I_0$ or $I_1$) to get a numerical or exact value for the integral

**递推降阶公式** — 将$I_n$（某函数的$n$次幂的积分）表示为$k < n$的$I_{n-k}$的递推关系，对于三角函数与双曲函数通常取$k=2$。

**例题:** 推导$I_n = \int_0^{\pi/2} \sin^n x dx$的降阶公式，并使用该公式计算$I_4$

1. 将被积函数拆分为两项以进行分部积分：
2. $$I_n = \int_0^{\pi/2} \sin^{n-1}x \times \sin x dx$$
3. 令$u = \sin^{n-1}x$，$dv = \sin x dx$，因此$du = (n-1)\sin^{n-2}x \cos x dx$，$v = -\cos x$
4. 应用分部积分法：
5. $$I_n = [-\sin^{n-1}x \cos x]_0^{\pi/2} + (n-1)\int_0^{\pi/2} \sin^{n-2}x \cos^2 x dx$$
6. 边界项的计算结果为0，代入$\cos^2 x = 1 - \sin^2 x$：
7. $$I_n = (n-1)\int_0^{\pi/2} \sin^{n-2}x (1 - \sin^2 x) dx = (n-1)I_{n-2} - (n-1)I_n$$
8. 整理得到递推关系：
9. $$I_n + (n-1)I_n = (n-1)I_{n-2} \implies nI_n = (n-1)I_{n-2}$$
10. 使用基础情况$I_0 = \int_0^{\pi/2} 1 dx = \pi/2$计算$I_4$：
11. $$I_4 = \frac{3}{4}I_2 = \frac{3}{4} \times \frac{1}{2}I_0 = \frac{3}{8} \times \frac{\pi}{2} = \frac{3\pi}{16}$$

> **考试提示**
>
> 在代入基础情况计算之前，务必明确写出递推关系，因为即使你最终计算出错，写出正确的递推关系也能获得相应分数。

*计算器:* allowed

## 计算直角坐标曲线与参数曲线的弧长

直角坐标曲线和参数曲线的弧长公式已收录在公式手册中。极坐标形式的弧长不属于FP3的考察范围。请务必先化简根号内的表达式，它通常可以化简为完全平方形式。

**例题:** 计算曲线$y = \frac{2}{3}x^{3/2}$在$x=0$到$x=3$之间的弧长

1. 将y对x求导：
2. $$\frac{dy}{dx} = x^{1/2} = \sqrt{x}$$
3. 代入直角坐标弧长公式：
4. $$s = \int_0^3 \sqrt{1 + (\frac{dy}{dx})^2} dx = \int_0^3 \sqrt{1 + x} dx$$
5. 计算该积分：
6. $$s = \left[\frac{2}{3}(1+x)^{3/2}\right]_0^3 = \frac{2}{3}(4^{3/2} - 1^{3/2}) = \frac{2}{3}(8 - 1) = \frac{14}{3} \approx 4.67$$

*计算器:* allowed

## 计算旋转体的表面积

绕x轴旋转得到的旋转体表面积公式为$S_x = 2\pi \int y ds$，其中$ds$是与曲线形式（直角坐标或参数形式）匹配的弧长微元。不要遗漏$y$项或$ds$项，这是常见的失分点。

**例题:** 求曲线$y = \sqrt{x}$（$0 \leq x \leq 4$）绕x轴旋转$2\pi$弧度后形成的旋转体表面积

1. 将y对x求导：
2. $$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$$
3. 写出直角坐标形式下的$ds$项：
4. $$ds = \sqrt{1 + (\frac{dy}{dx})^2} dx = \sqrt{1 + \frac{1}{4x}} dx = \sqrt{\frac{4x + 1}{4x}} dx = \frac{\sqrt{4x + 1}}{2\sqrt{x}} dx$$
5. 代入表面积公式：
6. $$S_x = 2\pi \int_0^4 y ds = 2\pi \int_0^4 \sqrt{x} \times \frac{\sqrt{4x + 1}}{2\sqrt{x}} dx$$
7. 化简被积函数，$\sqrt{x}$项可以抵消：
8. $$S_x = \pi \int_0^4 \sqrt{4x + 1} dx$$
9. 使用换元法$u = 4x + 1$，$du = 4dx$计算该积分：
10. $$S_x = \pi \times \frac{1}{4} \times \frac{2}{3}[(4x + 1)^{3/2}]_0^4 = \frac{\pi}{6}(17^{3/2} - 1) \approx 36.2$$

> **warning**
>
> 在计算表面积时绝对不能用$dx$代替$ds$，否则会得到错误结果。请务必代入与曲线形式对应的完整$ds$表达式。

*计算器:* allowed

## 常见错误

- **错误做法:** Using $x = a \sinh u$ substitution for $\sqrt{x^2 - a^2}$ instead of $x = a \cosh u$
  - 原因: $\sinh^2 u + 1 = \cosh^2 u$, so $\sqrt{a^2 \sinh^2 u + a^2} = a \cosh u$, which works for $\sqrt{x^2 + a^2}$ not $\sqrt{x^2 - a^2}$
  - 正确做法: Use $x = a \cosh u$ for $\sqrt{x^2 - a^2}$, as $\cosh^2 u - 1 = \sinh^2 u$, simplifying the surd to $a \sinh u$ for $x \geq a$
- **错误做法:** Writing $S_x = 2\pi \int y dx$ instead of $2\pi \int y ds$ for surface area calculations
  - 原因: Surface area depends on the length of the curve segment, not just the x-axis interval, so omitting $ds$ gives an incorrect value
  - 正确做法: Always substitute the correct $ds$ form (Cartesian or parametric) matching the curve's equation before evaluating the integral
- **错误做法:** Splitting $I_n = \int \sin^n x dx$ as $u = \sin^n x$, $dv = dx$ when deriving reduction formulae
  - 原因: This split leads to an integral with $x \sin^{n-1}x \cos x$, which is more complex and does not produce a lower power of $\sin x$
  - 正确做法: Split powers of trigonometric/hyperbolic functions into a $(n-1)$ power term and a 1 power term for integration by parts to get a recurrence relation with $I_{n-2}$
- **错误做法:** Trying to integrate inverse functions directly instead of using integration by parts
  - 原因: Inverse functions have no elementary antiderivative when integrated directly, so this approach will not work
  - 正确做法: For integrals of inverse functions like $\int \arctan x dx$, set $u = $ inverse function, $dv = dx$, then evaluate the resulting simpler integral
- **错误做法:** Using incorrect limits for parametric arc length/surface area integrals
  - 原因: The parameter limits must correspond exactly to the start and end points of the curve segment being measured, not the x-axis limits
  - 正确做法: Check that you are using the correct parameter interval that maps to the curve endpoints given in the question

## 速查表

| Concept | Formula / Technique | Exam Tip |
| --- | --- | --- |
| Integrate inverse functions | Use by parts: $u = f^{-1}(x)$, $dv = dx$ | Use formula book surd results directly for the resulting integral to save time |
| $\sqrt{a^2 + x^2}$ substitution | $x = a \sinh u$ | Simplifies to $a \cosh u$, result maps to $\text{arsinh}(x/a)$ |
| $\sqrt{a^2 - x^2}$ substitution | $x = a \sin \theta$ | Simplifies to $a \cos \theta$, result maps to $\arcsin(x/a)$ |
| $\sqrt{x^2 - a^2}$ substitution | $x = a \cosh u$ | Simplifies to $a \sinh u$, result maps to $\text{arcosh}(x/a)$ |
| Reduction formula derivation | Use by parts: $u = f(x)^{n-1}$, $dv = f(x)dx$ | State the recurrence relation explicitly before applying to base cases |
| Cartesian arc length | $s = \int \sqrt{1 + (dy/dx)^2} dx$ | Simplify the term inside the square root first, look for perfect squares |
| Parametric arc length | $s = \int \sqrt{(dx/dt)^2 + (dy/dt)^2} dt$ | Use parameter limits matching the curve endpoints, not x/y limits |
| Surface area (x-axis) | $S_x = 2\pi \int y ds$ | Use the $ds$ form matching the curve's equation (Cartesian/parametric) |

## 下一步

Now that you have mastered FP3 integration techniques, you are ready to apply these skills to the remaining core FP3 topics, as well as synoptic questions across the Further Mathematics specification. Integration of hyperbolic functions and reduction formulae are often combined with differential equations topics in FP4, while arc length and surface area calculations may appear in synoptic questions linking to coordinate geometry in FP2. Be sure to practice full past paper questions for this topic, as exam questions typically combine multiple integration techniques in a single multi-part problem, worth 8-12 marks. You should also revise integration techniques from Pure 3 to ensure you can quickly recall substitution and by parts rules under timed exam conditions.

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