Exercise 20F.1: Mastering Integration by Substitution | 练习20F.1：掌握换元积分法

📚 Exercise 20F.1: Mastering Integration by Substitution | 练习20F.1：掌握换元积分法

Integration by substitution is one of the most powerful techniques in calculus, allowing us to transform complicated integrals into simpler, standard forms. In the specific context of IB Mathematics, Exercise 20F.1 presents a carefully graded set of problems designed to build fluency in recognising when and how to apply this method. This article provides a thorough walkthrough of the concepts, procedures, and typical pitfalls associated with substitution, serving as your complete companion to the exercise.

换元积分法是微积分中最强大的技巧之一，能让我们将复杂的积分转化为更简单、标准的形式。在IB数学的具体背景下，练习20F.1提供了一套精心设计的题目，旨在培养大家识别何时以及如何运用这种方法的熟练度。本文将深入讲解换元法的概念、步骤和常见误区，让你在完成练习时得心应手。

1. The Core Idea of Substitution | 换元法的核心思想

n rule for differentiation. When you differentiate a composite function f(g(x)), you get f'(g(x)) * g' pse the whole expression into an integral solely in terms of u and du, which is much easier to evaluate.

换元积分法本质上是微分的链式法则的逆运算。当你对复合函数 f(g(x)) 求导时，会得到 f'(g(x)) * g'(x) 。在积分中，如果被积函数类似于这种结构，通过精心选择代换变量 u = g(x)，就能将整个表达式简化为只关于 u 和 du 的积分，从而轻松求解。

2x dx, transforming the integral into ∫ cos u du, an instantly recognisable standard form.

例如，看到积分 ∫ 2x·cos(x²) dx，我们会发现 x² 的导数 2x 恰好就在余弦函数旁边。令 u = x²，则 du = 2x dx，积分就变成了 ∫ cos u du，一个一眼就能认出的标准形式。

2. The Reverse Chain Rule in Practice | 链式法则逆向的实际应用

The formal statement of integration by substitution is: ∫ f(g(x))·g' ntegrand must contain both the composite outer function f(g(x)) and the inner derivative g'(x) (up to a constant factor) for the substitution to work smoothly.

换元积分法的正式表述为：∫ f(g(x))·g'(x) dx = ∫ f(u) du，其中 u = g(x)。这一公式强调，被积函数必须同时包含复合的外层函数 f(g(x)) 和内层导数 g'(x)（或至多差一个常数因子），代换才能顺利奏效。

In Exercise 20F.1, many problems are crafted specifically to help you spot this derivative–function pairing. You will frequently see integrals of the form ∫ (something)ⁿ × (its derivative) dx, ∫ e^something e) dx. The sooner you can identify these patterns, the faster you will solve the problems.

在练习20F.1中，很多题目正是为了帮你发现这种导数与函数的配对而设计的。你会频繁见到 ∫ (某式)ⁿ × (其导数) dx、∫ e^某式 × (其导数) dx 或 ∫ cos(某式) × (其导数) dx 等形式。越早识别这些模式，解题速度就越快。

3. Choosing the Right Substitution | 选择合适的代换

A common question from students is 'How do I know what to set as u?' The answer lies in looking for an inner function whose derivative also appears as a factor. Key candidates for u often include: expressions inside a bracket raised to a power, exponents of e, denominators of fractions, and arguments of trigonometric or logarithmic functions.

学生常问怎么知道该把什么设为 u，答案在于寻找那个内层函数，且其导数也恰好作为因子出现。常见的 u 候选通常包括：幂的括号内表达式、e 的指数、分数的分母，以及三角函数或对数函数的自变量。

This adjustment by a constant factor is perfectly legal and very common in Exercise 20F.1.

有时导数并不精确出现，只差一个常数倍。例如 ∫ x·sin(3x²) dx，3x² 的导数是 6x，但只有 x。你仍然可以设 u = 3x²，那么 du = 6x dx，所以 x dx = (1/6) du，积分变为 (1/6) ∫ sin u du。这种用常数因子进行调整的方法完全合理，并且在练习20F.1中非常普遍。

4. Step-by-Step Procedure for Substitution | 换元法的逐步步骤

errors. We recommend the following sequence. First, identify u by looking at the integrand's inner structure. Write u = g(x). Next, compute du/dx, then rewrite to isolate dx: du = g'(x) dx, so dx = du / g' s the antiderivative in the original variable. If you are dealing with a definite integral, you can either change the limits to u-values or substitute back before applying the original limits.

采用结构化的方法可以消除随意猜测并减少错误。我们建议以下顺序：首先观察被积函数的内层结构以确定 u，写出 u = g(x)。接着计算 du/dx，然后分离出 dx：du = g'(x) dx，从而 dx = du / g'(x)。将所有的 x 和 dx 替换为它们关于 u 的等价形式，确保被积函数完全变成 u 的函数。然后对 u 求积分。最后代回 u = g(x)，用原变量表示原函数。如果处理的是定积分，可以将积分限转换成 u 值，或者在代回原变量后再代入原始上下限。

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5. Worked Example 1: ∫ x√x²+1 dx | 例题1：∫ x√x²+1 dx

Then du/dx = 2x, so du = 2x dx, which gives x dx = (1/2) du. The integral becomes (1/2) ∫ √u du, which is (1/2) ∫ u^1/2 du. Integrating gives (1/2) * (2/3) u^3/2 + C = (1/3) u^3/2 + C. Finally, substitute back: (1/3)(x²+1)^3/2 + C.

我们设 u = x² + 1，因为导数 2x 与根号外的 x 因子紧密相关。于是 du/dx = 2x，得到 du = 2x dx，即 x dx = (1/2) du。积分变为 (1/2) ∫ √u du，也就是 (1/2) ∫ u^1/2 du。积分得 (1/2) * (2/3) u^3/2 + C = (1/3) u^3/2 + C。最后代回：(1/3)(x²+1)^3/2 + C。

6. Worked Example 2: ∫ sin²x cos x dx | 例题2：∫ sin²x cos x dx

es to (1/3) u³ + C. Substituting back yields (1/3) sin³x + C. Notice how the integrand immediately simplifies into a basic power function once the substitution is made.

这里明确的选择是 u = sin x，因为其导数 cos x 出现在被积函数中。于是 du = cos x dx。积分变为 ∫ u² du，积分得 (1/3) u³ + C。代回后得到 (1/3) sin³x + C。注意一旦进行代换，被积函数立即简化为基本的幂函数。

7. Worked Example 3: Definite Integral ∫ from 0 to 1 of xe^x² dx | 例题3：定积分 ∫₀¹ xe^x² dx

2) du. When x = 0, u = 0; when x = 1, u = 1. The integral becomes (1/2) ∫ from 0 to 1 of e^u du = (1/2) [e^u trates the convenience of converting the limits to u; there is no need to return to x before substituting the limits.

设 u = x²，则 du = 2x dx，因此 x dx = (1/2) du。当 x = 0 时，u = 0；当 x = 1 时，u = 1。积分变为 (1/2) ∫₀¹ e^u du = (1/2) [e^u]₀¹ = (1/2)(e - 1)。这显示了把积分限转换为 u 的方便之处，无需先变回 x 再代入上下限。

8. Common Pitfalls and How to Avoid Them | 常见误区及避免方法

integrand has only dx, you must introduce (1/2) accordingly. Finally, when handling definite integrals, ensure you change the limits to match the substitution variable, or clearly back-substitute before evaluating.

一个常见错误是忘记将 dx 完全替换为 du。有些同学代换了 u，但积分中仍保留 dx，导致两个变量混在一起。务必在继续之前用 du 表达 dx。另一个错误是忽略常数因子的调整：若 du = 2 dx 而被积函数只有 dx，必须相应地乘以 (1/2)。最后，处理定积分时，确保将积分限改成与代换变量匹配，或者在代入原限前清楚地进行回代。

) dx cannot easily be handled by substitution because there is no x factor to pair with the derivative of x². Recognising when substitution is not applicable saves valuable time in IB exams.

另一个不易察觉的陷阱是对不适合代换模式的积分强行使用换元法。例如 ∫ sin(x²) dx 无法通过代换轻易求解，因为没有 x 因子与 x² 的导数配对。在IB考试中，识别出代换法不可用能节省宝贵时间。

9. Practice Strategy for Exercise 20F.1 | 练习20F.1的做题策略

Start by scanning each integral for a clear 'inner function' whose derivative is present alongside. If the derivative appears exactly, the problem is straightforward. If a constant adjustment is needed, write down du and the compensating factor explicitly. Work through the exercise in order, as the problems usually increase in difficulty, building from direct substitution to applying trigonometric identities before substituting.

首先快速扫描每个积分，寻找清晰的“内层函数”，且其导数一同出现。如果导数精确存在，题目就很简单。如果需要常数调整，明确写下 du 和相应的补偿因子。按顺序完成练习，因为题目通常由易到难，从直接代换逐步过渡到先应用三角恒等式再代换。

After completing the exercise, revisit the problems that required the most effort and try to identify the substitution on sight. This will build your intuition and speed for examination conditions. Also, practise both indefinite and definite forms, and get comfortable with changing limits.

完成练习后，重温那些最费劲的题目，并尝试一眼识别出代换。这会增强你的直觉和应考速度。此外，不定积分和定积分的形式都要练习，并熟悉积分限的转换。

10. Summary and Key Takeaways | 总结与要点

Integration by substitution transforms a complicated integral into a simpler one by replacing a component with a new variable u. Mastery relies on recognising the derivative of your chosen u within the integrand. Always double-check that the integral is entirely in u before integrating, and remember to back-substitute or change limits for definite integrals. Exercise 20F.1 provides an essential foundation in this technique, and careful, step-by-step practice will make substitution a natural tool in your calculus toolkit.

换元积分法通过将组成部分替换为新变量 u，把复杂的积分转化为简单的积分。掌握它的关键在于识别出被积函数中存在所选 u 的导数。在积分前，务必反复检查积分是否完全以 u 表示，并记住要回代原变量或改变定积分限。练习20F.1为这项技巧提供了必要的基础，通过仔细、逐步的练习，换元法将成为你微积分工具箱中运用自如的工具。

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