Exercise 21E.1: Integration by Substitution | 练习21E.1：代换积分法

📚 Exercise 21E.1: Integration by Substitution | 练习21E.1：代换积分法

Welcome to this focused revision guide for Exercise 21E.1, a key set of problems from the IB Mathematics Analysis & Approaches syllabus. This exercise builds your skills in one of the most powerful integration techniques: substitution. Whether you are preparing for a test or trying to master the reverse chain rule, this article will walk you through the essential concepts, worked examples, and common mistakes so you can approach each integral with confidence.

欢迎阅读这篇针对练习21E.1的专项复习指南，该练习出自IB数学分析与方法课程大纲。这一练习帮助你培养掌握最强大的积分技巧之一：代换积分法。无论你是在准备考试还是努力掌握反向链式法则，本文都将带你梳理基本概念、例题演示和常见错误，让你能够自信地应对每一个积分。

1. Understanding Integration by Substitution | 理解代换积分法

Integration by substitution is a technique used to simplify integrals by changing the variable of integration. It is essentially the reverse process of the chain rule for differentiation. Instead of trying to integrate a complicated expression directly, we introduce a new variable, typically denoted u, to rewrite the integral into a standard form that we can easily evaluate.

代换积分法是一种通过更换积分变量来简化积分的技巧。它本质上是微分链式法则的逆过程。我们不是直接尝试积分一个复杂的表达式，而是引入一个新变量，通常记作 u，将积分改写成一个易于计算的标准形式。

This method is particularly useful when the integrand contains a function and its derivative, or when a composition of functions makes the integral look intimidating. Once you become comfortable with choosing the right substitution, a wide class of integrals becomes manageable.

当被积函数包含一个函数与其导数，或者函数复合使得积分看起来令人生畏时，这个方法尤其有用。一旦你熟悉了如何选择合适的代换，一大类积分都将变得易于处理。

2. The Core Idea: Reverse Chain Rule | 核心思想：链式法则的逆运算

Recall the chain rule for differentiation: if y = f(u) and u = g(x), then dy/dx = f'(u) · g'(x). The integration equivalent says that if we can spot an integrand of the form f'(g(x)) · g'(x), then its antiderivative is simply f(g(x)) + C. By letting u = g(x), we get du/dx = g'(x), so du = g'(x) dx, and the integral transforms into ∫ f'(u) du = f(u) + C.

回顾微分的链式法则：如果 y = f(u) 且 u = g(x)，那么 dy/dx = f'(u) · g'(x)。相应的积分形式告诉我们，如果能将被积函数识别为 f'(g(x)) · g'(x) 的形式，它的原函数就是 f(g(x)) + C。令 u = g(x)，我们得到 du/dx = g'(x)，因此 du = g'(x) dx，积分就转化为 ∫ f'(u) du = f(u) + C。

This is the heartbeat of substitution: we deliberately choose u to absorb the inner function, and the derivative g'(x) dx is replaced by du, provided it appears (perhaps up to a constant factor) in the original integral. Exercise 21E.1 will repeatedly test your ability to match patterns to this reverse chain rule structure.

这就是代换积分法的核心：我们有意识地选择 u 来吸收内层函数，而导数 g'(x) dx 被 du 替换，前提是它（可能相差一个常数倍）出现在原积分中。练习21E.1将反复检验你将这些模式与反向链式法则结构相匹配的能力。

3. Step-by-Step Procedure for Substitution | 代换积分法的逐步步骤

To apply integration by substitution effectively, follow a systematic routine. Even when the integral looks messy, these steps will keep you on track:

要高效应用代换积分法，请遵循一套系统的流程。即使积分看起来很混乱，这些步骤也能让你保持正确的方向：

Step 1: Identify a suitable inner function g(x) to set as u. Look for a function whose derivative g'(x) (or a constant multiple of it) appears elsewhere in the integrand.
步骤1： 识别一个合适的内层函数 g(x) 并将其设为 u。寻找其导数 g'(x)（或其常数倍）出现在被积函数中的其他位置。
Step 2: Compute du/dx and isolate dx (or directly write du = g'(x) dx).
步骤2： 计算 du/dx 并解出 dx（或直接写出 du = g'(x) dx）。
Step 3: Substitute u and du into the integral, replacing all x-expressions and dx. The new integral should contain only the variable u.
步骤3： 将 u 和 du 代入积分，替换所有含 x 的表达式和 dx。新的积分应只包含变量 u。
Step 4: Evaluate the simplified integral with respect to u.
步骤4： 计算简化后的关于 u 的积分。
Step 5: Substitute back u = g(x) to rewrite the antiderivative in terms of the original variable x. Add the constant of integration + C.
步骤5： 将 u = g(x) 代回，用原变量 x 写出原函数。加上积分常数 + C。

For definite integrals, you can either change the limits of integration to u-values or substitute back before evaluating the limits—both methods appear in Exercise 21E.1.

对于定积分，你可以将积分上下限换成对应的 u 值，也可以先代回再代入原上下限——这两种方法在练习21E.1中都会出现。

4. Choosing the Right ‘u’ | 选择合适的“u”

The most challenging part of substitution is often deciding what to let u equal. A good rule of thumb is to set u as the inner part of a composite function, especially the part inside a power, a trigonometric function, an exponential, or a logarithm. For instance, if you see an expression like (2x³ + 1)⁵, try u = 2x³ + 1; then du = 6x² dx, and the x² dx from the integrand will help replace dx with du/6x² after adjusting constants.

代换积分法最具挑战性的部分通常是决定让 u 等于什么。一条好用的经验法则是将 u 设为复合函数的内层部分，特别是幂、三角函数、指数或对数内部的部分。例如，如果你看到像 (2x³ + 1)⁵ 这样的表达式，可以尝试令 u = 2x³ + 1；那么 du = 6x² dx，而被积函数中的 x² dx 部分经过常数调整后将帮你用 du 替换 dx。

Another tip: do not be afraid to try a substitution and find it does not simplify the integral nicely. With practice, you will develop an instinct for the patterns that appear in standard IB problems. Exercise 21E.1 is designed to train exactly this pattern recognition.

另一个提示：不必害怕尝试某个代换后发现它并不能很好地简化积分。通过练习，你会对标准IB问题中出现的模式培养出直觉。练习21E.1正是为了训练这种模式识别能力而设计的。

5. Example 1: Polynomial Substitution | 例题1：多项式代换

Let’s work through an integral similar to those you will meet in Exercise 21E.1. Consider ∫ 3x² (2x³ + 1)⁴ dx. The derivative of the inner function 2x³ + 1 is 6x², which is almost the 3x² sitting outside. This prompts the substitution u = 2x³ + 1.

让我们演练一个与练习21E.1中类似的积分。考虑 ∫ 3x² (2x³ + 1)⁴ dx。内层函数 2x³ + 1 的导数是 6x²，这和外部的 3x² 几乎一致。这提示我们采用代换 u = 2x³ + 1。

Then du/dx = 6x², so du = 6x² dx. Notice our integrand has 3x² dx, which is exactly ½ du. The integral becomes ∫ (2x³+1)⁴ · (3x² dx) = ∫ u⁴ · (½ du) = ½ ∫ u⁴ du. Now integrate: ½ · (u⁵/5) = u⁵/10. Finally, substitute back: (2x³ + 1)⁵ / 10 + C.

那么 du/dx = 6x²，所以 du = 6x² dx。注意到我们的被积函数中有 3x² dx，这恰好是 ½ du。积分变成 ∫ (2x³+1)⁴ · (3x² dx) = ∫ u⁴ · (½ du) = ½ ∫ u⁴ du。现在积分：½ · (u⁵/5) = u⁵/10。最后代回：(2x³ + 1)⁵ / 10 + C。

∫ 3x² (2x³ + 1)⁴ dx = (2x³ + 1)⁵ / 10 + C

6. Example 2: Trigonometric Substitution | 例题2：三角代换

Trigonometric integrals are another common visitor in Exercise 21E.1. Take ∫ sin³ x cos x dx. We spot that the derivative of sin x is cos x, so letting u = sin x works perfectly. Then du = cos x dx, and the sin³ x becomes u³.

三角函数的积分也是练习21E.1中的常客。以 ∫ sin³ x cos x dx 为例。我们发现 sin x 的导数是 cos x，所以令 u = sin x 非常合适。那么 du = cos x dx，而 sin³ x 变为 u³。

The integral simplifies to ∫ u³ du = u⁴/4 + C. Replacing u with sin x gives the final answer: (1/4) sin⁴ x + C. Notice how the cos x dx combination was directly absorbed into du—this is the classic reverse chain rule pattern.

积分简化为 ∫ u³ du = u⁴/4 + C。将 u 替换为 sin x 得到最终答案：(1/4) sin⁴ x + C。注意 cos x dx 这个组合是如何被 du 直接吸收的——这就是经典的反向链式法则模式。

∫ sin³ x cos x dx = (1/4) sin⁴ x + C

7. Handling Definite Integrals with Substitution | 用代换法处理定积分

When a definite integral is involved, such as ∫ₐᵇ f(g(x)) g'(x) dx, we have two options. The more elegant method is to change the limits as well: once u = g(x), the lower limit x = a becomes u = g(a), and the upper limit x = b becomes u = g(b). Then evaluate the u-integral directly, never returning to x.

当涉及定积分时，例如 ∫ₐᵇ f(g(x)) g'(x) dx，我们有两种处理方式。更优雅的方法是同时更换积分上下限：一旦设定 u = g(x)，下限 x = a 变成 u = g(a)，上限 x = b 变成 u = g(b)。然后直接计算关于 u 的积分，不再回到 x。

For example, evaluate ∫₀¹ 2x (x² + 1)³ dx. Let u = x² + 1. Then du = 2x dx. When x = 0, u = 1; when x = 1, u = 2. The integral becomes ∫₁² u³ du = [u⁴/4]₁² = 16/4 – 1/4 = 15/4. Alternatively, you can integrate indefinitely, substitute back, and then apply the original limits—both yield the same result, but changing limits often saves time.

例如，计算 ∫₀¹ 2x (x² + 1)³ dx。令 u = x² + 1。那么 du = 2x dx。当 x = 0 时，u = 1；当 x = 1 时，u = 2。积分变为 ∫₁² u³ du = [u⁴/4]₁² = 16/4 – 1/4 = 15/4。或者，你也可以先计算不定积分，代回后代入原上下限——两种方法结果相同，但更换上限通常更节省时间。

In Exercise 21E.1, you will be encouraged to practice both approaches to see which feels more natural for different problems.

在练习21E.1中，你将需要练习这两种方法，看看在处理不同问题时哪种感觉更自然。

8. Common Pitfalls and How to Avoid Them | 常见陷阱及如何避免

Substitution is straightforward in theory, but small mistakes can lead to wrong answers. One frequent error is forgetting to replace dx completely. Students sometimes mix x and u in the same integral, which is invalid. Always ensure that after substitution, no x terms remain—everything must be expressed in u and du.

代换积分法在理论上很简单，但小错误却可能导致错误答案。一个常见的失误是忘记完全替换 dx。学生有时会在同一个积分中混杂 x 和 u，这是不允许的。始终确保代换之后没有 x 项残留——所有内容都必须用 u 和 du 表达。

Another trap is mishandling constants. If du = 2 dx but your integrand only has dx, you must write dx = ½ du. Forgetting that constant factor will throw off the whole evaluation. Also, with definite integrals, failing to update the limits is a classic mistake—always check that your limits match the variable of integration.

另一个陷阱是处理常数不当。如果 du = 2 dx 但你的被积函数中只有 dx，你必须写出 dx = ½ du。忘记那个常数因子会搞乱整个计算。此外，对于定积分，不更新上下限是一个典型错误——始终确保你的上下限与积分变量相匹配。

Finally, when substituting back, do not forget to add the constant + C for indefinite integrals. IB examiners are keen to deduct marks for missing constants of integration in long-answer questions.

最后，在代回时，不要忘记对不定积分加上常数 + C。IB考官很乐意在长题中因缺少积分常数而扣分。

9. Practice Strategy for Exercise 21E.1 | 练习21E.1的练习策略

To get the most out of Exercise 21E.1, avoid rushing through the problems mechanically. Start by scanning each integral and identifying the likely ‘u’ before picking up your pen. Say aloud why you think a particular substitution will work—this verbalisation strengthens your pattern recognition.

为了最大化练习21E.1的收获，不要机械地快速刷题。动笔之前先浏览每个积分并识别可能的“u”。大声说出你为什么认为某个代换会奏效——这种口头表达会强化你的模式识别能力。

Work through a mix of indefinite and definite integrals, and for the definite ones, try solving each twice: once by changing limits, once by back-substituting. Compare your answers to ensure consistency. Also, after you obtain a solution, differentiate it to verify that you retrieve the original integrand—this self-check is invaluable for building accuracy.

交替练习不定积分和定积分，对于定积分，尝试用两种方法各解一遍：一次通过更换上下限，一次通过代回。比较答案以确保一致性。此外，得到解答后，对其进行求导，验证是否能回到原始的被积函数——这种自我检查对提高准确性非常宝贵。

If you get stuck, revisit the step-by-step framework in Section 3. Often, simply writing down u, du, and the transformed integral on scratch paper reveals the path forward.

如果你卡住了，重新审视第3节中的逐步框架。通常，只需在草稿纸上写下 u、du 和变换后的积分，前进的道路就会显现出来。

10. Conclusion and Next Steps | 结论与下一步

Integration by substitution is a cornerstone of IB Mathematics, and Exercise 21E.1 is your training ground. By mastering the selection of u, carefully handling differentials, and practicing both indefinite and definite cases, you are equipping yourself with a tool that will appear repeatedly in exams, from polynomial integrals to trigonometric and exponential forms.

代换积分法是IB数学的基石，而练习21E.1就是你的训练场。通过掌握 u 的选择、仔细处理微分、并练习不定和定积分两种情况，你正在为自己配备一个在考试中反复出现的工具，从多项式积分到三角和指数形式都离不开它。

Once you finish Exercise 21E.1, challenge yourself with more complex integrals that require algebraic manipulation before substitution, or those that involve inverse trigonometric functions. The confidence you build here will also support later topics like integration by parts and differential equations.

完成练习21E.1后，可以挑战一些在代换前需要代数变形的更复杂积分，或者涉及反三角函数的积分。你在此处建立的信心也会支撑后续学习，如分部积分法和微分方程。

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Exercise 21E.1: Integration by Substitution | 练习21E.1：代换积分法