📚 Mastering Integration by Substitution: Exercise 20F.2 | 掌握换元积分法：习题20F.2详解
Integration by substitution is one of the most versatile techniques in the IB Mathematics HL syllabus. This exercise set, 20F.2, is designed to build the pattern-recognition skills and procedural fluency students need to tackle both indefinite and definite integrals confidently. In this article, we work through the core ideas, demonstrate several fully solved examples, highlight common mistakes, and provide additional practice problems from the same set.
换元积分法是 IB 数学高阶课程中最灵活的技巧之一。习题 20F.2 旨在培养学生识别积分模式的能力和熟练的运算步骤，使他们充满信心地应对不定积分与定积分。本文将通过核心概念讲解、若干完整例题、常见错误分析和额外练习题，带你全面掌握这一方法。
1. Overview of Exercise 20F.2 | 习题20F.2概览
oal is not merely to find the answer but to become adept at selecting the function u = g(x) that makes the integrand 'collapse' neatly.
习题 20F.2 包含了多项式、三角函数、指数、对数以及有理函数等被积函数的综合题型。每一个题目都经过精心设计，使得一个恰当的代换能将原积分转化为基本积分形式，如幂函数、指数或对数积分。练习的目标不仅仅是求出答案，而是熟练选择让被积函数“简化”的函数 u = g(x)。
2. The Core Principle of Substitution | 换元法的核心原理
n reverses the chain rule. If we recognise that an integrand can be expressed as f(g(x))·g'(x), we let u = g(x) so that du = g'(x) dx. The integral then reduces to ∫ f(u) du, which is often straightforward.
换元法本质上是链式法则的逆运算。如果我们识别出被积函数可以写成 f(g(x))·g'(x) 的形式，便可以令 u = g(x)，从而有 du = g'(x) dx。此时积分简化为 ∫ f(u) du，通常很容易计算。
∫ f(g(x))·g'(x) dx = ∫ f(u) du, where u = g(x)
This seemingly simple statement underpins all the problems in Exercise 20F.2. The key challenge is training your eye to spot the inner function g(x) and confirm that its derivative (or a constant multiple) appears as a factor in the integrand.
这个看似简单的公式是习题 20F.2 所有问题的基础。关键挑战在于训练你的眼睛，能够迅速发现内层函数 g(x)，并确认其导数（或常数倍）是否作为被积函数的一个因子出现。
3. Worked Example 1: Power Rule Substitution | 例题1：幂函数型代换
Problem: Evaluate ∫ 2x (x² + 1)⁵ dx.
题目：计算 ∫ 2x (x² + 1)⁵ dx。
ion x² + 1 is 2x, which appears exactly in the integrand. Let u = x² + 1. Then du = 2x dx.
步骤1：注意到复合函数 (x² + 1)⁵ 及其导数的线索。内层函数 x² + 1 的导数为 2x，恰好出现在被积函数中。令 u = x² + 1，则 du = 2x dx。
Step 2: Substitute everything: ∫ (x² + 1)⁵ · 2x dx = ∫ u⁵ du.
步骤2：代入所有项：∫ (x² + 1)⁵ · 2x dx = ∫ u⁵ du。
Step 3: Integrate with respect to u: ∫ u⁵ du = u⁶/6 + C.
步骤3：对 u 积分：∫ u⁵ du = u⁶/6 + C。
Step 4: Replace u with x² + 1 to return to the original variable: (x² + 1)⁶/6 + C.
步骤4：将 u 代回 x² + 1，得到原变量的表达式：(x² + 1)⁶/6 + C。
Answer: (x² + 1)⁶/6 + C
This example illustrates the most common pattern: a linear inner function raised to a power, multiplied by its derivative.
这个例子展示了最常见的模式：内层线性函数的幂，乘以其导数。
4. Worked Example 2: Trigonometric Substitution | 例题2：三角函数代换
Problem: Find ∫ sin³ θ cos θ dθ.
题目：求 ∫ sin³ θ cos θ dθ。
= sin θ, giving du = cos θ dθ. The integral becomes ∫ u³ du = u⁴/4 + C = (1/4) sin⁴ θ + C.
这里被积函数包含 sin³ θ 和 cos θ。由于 sin θ 的导数是 cos θ，这提示我们可以设 u = sin θ，从而 du = cos θ dθ。积分变为 ∫ u³ du = u⁴/4 + C = (1/4) sin⁴ θ + C。
Always verify that the derivative matches exactly or can be matched by a constant factor. In this case the match is exact, making the substitution extremely efficient.
始终要检查导数是否精确匹配，或者能否通过常数因子调整后匹配。此处精确匹配，代换极为高效。
5. Worked Example 3: Logarithmic Substitution | 例题3：对数型代换
Problem: Evaluate ∫ (ln x)/x dx.
题目：计算 ∫ (ln x)/x dx。
r. Set u = ln x, so du = (1/x) dx. The integral becomes ∫ u du = u²/2 + C = (ln x)²/2 + C.
注意到 ln x 的导数是 1/x，恰好出现在分子中。设 u = ln x，则 du = (1/x) dx。积分变成 ∫ u du = u²/2 + C = (ln x)²/2 + C。
This pattern appears frequently in IB examinations, especially when the integrand is a quotient involving natural logarithms.
这种模式在 IB 考试中经常出现，尤其是被积函数为含有自然对数的分式时。
6. Handling Definite Integrals | 处理定积分
o a definite integral, you must transform the limits of integration. Suppose ∫ₐᵇ f(g(x))·g' u = g(x), when x = a, u = g(a); when x = b, u = g(b). The definite integral becomes ∫_g(a)g(b) f(u) du, and there is no need to revert to x.
当对定积分使用换元法时，必须同时变换积分上下限。对于 ∫ₐᵇ f(g(x))·g'(x) dx，令 u = g(x)，当 x = a 时 u = g(a)；x = b 时 u = g(b)。定积分变为 ∫_g(a)g(b) f(u) du，无需再切换回 x。
Example: Evaluate ∫₀¹ x/(1 + x²) dx.
例题：计算 ∫₀¹ x/(1 + x²) dx。
ts: when x = 0, u = 1; when x = 1, u = 2. The integral becomes (1/2) ∫₁² 1/u du = (1/2)[ln|u|]₁² = (1/2)(ln 2 – ln 1) = (1/2) ln 2.
设 u = 1 + x²，则 du = 2x dx，因此 x dx = (1/2) du。变换积分限：x = 0 时 u = 1；x = 1 时 u = 2。积分变为 (1/2) ∫₁² 1/u du = (1/2)[ln|u|]₁² = (1/2)(ln 2 – ln 1) = (1/2) ln 2。
Forgetting to update the limits is one of the most frequent errors students make in exams. Always pause and explicitly write the new u-values before integrating.
忘记更新积分限是学生在考试中最常犯的错误之一。在积分之前务必停下来，明确写出新的 u 值。
7. Common Pitfalls and How to Avoid Them | 常见陷阱与对策
Even a strong understanding of substitution can be undermined by small mistakes. Here are the most common pitfalls seen in Exercise 20F.2 tasks:
即便对换元法有扎实的理解，一些细小的错误也可能使答案前功尽弃。以下是习题 20F.2 中常见的陷阱：
- Misidentifying the inner function: Not every composite function has its derivative present. Sometimes you need to rearrange constants.
错误识别内层函数：并非每个复合函数都恰好有对应的导数，有时需要调整常数。 - Forgetting to substitute dx entirely: Writing du = dx or omitting dx leads to an incorrect integral.
忘记彻底替换 dx：错误地写 du = dx 或遗漏 dx 都会导致错误积分。 - Leaving the answer in terms of u for indefinite integrals: Must revert to the original variable.
不定积分最终答案仍用 u 表示：必须换回原变量。 - In definite integrals, failing to change the limits: Produces a numerically incorrect result.
定积分中未能改变积分限：得到错误数值。
To avoid these, always write down u, du/dx, solve for dx, and update limits if applicable. Double-check that the derivative appears, perhaps with a constant factor that can be adjusted outside the integral.
为了避免这些问题，务必逐一写出 u、du/dx、解出 dx，并在必要时更新积分限。再次确认导数确实以某个因子出现，可能只需在积分号外调整常数。
8. When to Use Substitution: Recognition Patterns | 何时使用换元法：识别模式
Developing pattern recognition is central to mastering Exercise 20F.2. Look for these signs:
培养模式识别能力是掌握习题 20F.2 的核心。请留意以下信号：
- ised to a power, multiplied by its derivative (or constant multiple) → power substitution.
一个函数的幂乘以它的导数（或常数倍）→ 幂代换。 - ric functions where one function is the derivative of another → u = sin θ, u = cos θ, etc.
三角函数中一个函数恰好是另一个的导数 → 设 u = sin θ, u = cos θ 等。 - An integrand of the form f'(x)/f(x) → u = f(x) gives ln|f(x)| directly.
形如 f'(x)/f(x) 的被积函数 → 设 u = f(x) 直接得到 ln|f(x)|。 - Exponential integrals with x·ex² → u = x².
含 x·ex² 的指数积分 → 设 u = x²。
ax + b to simplify the integrand. The problems in 20F.2 gradually build up this instinct.
如果不能立即发现导数，尝试简单的线性代换如 u = ax + b 来简化被积函数。20F.2 中的问题正是逐步培养这种直觉。
9. Practice Problems from Exercise 20F.2 | 习题20F.2练习精讲
Let us apply these ideas to a few more problems typical of the set. Each one reinforces a specific substitution flavour.
让我们将这些思路应用于该练习中另外几个典型问题，每个题目都强化一种特定的代换方式。
(a) ∫ x ex² dx — Set u = x², du = 2x dx ⇒ (1/2) ∫ eu du = (1/2) ex² + C.
(a) ∫ x ex² dx — 设 u = x²，du = 2x dx ⇒ (1/2) ∫ eu du = (1/2) ex² + C。
(b) ∫ 1/(x ln x) dx — Let u = ln x, du = (1/x) dx ⇒ ∫ 1/u du = ln|ln x| + C. Note the required condition x > 0, x ≠ 1.
(b) ∫ 1/(x ln x) dx — 令 u = ln x，du = (1/x) dx ⇒ ∫ 1/u du = ln|ln x| + C。注意需满足 x > 0 且 x ≠ 1。
(c) ∫ √2x + 1 dx — Use u = 2x + 1, du = 2 dx ⇒ (1/2) ∫ √u du = (1/2)·(2/3) u3/2 + C = (1/3)(2x + 1)3/2 + C.
(c) ∫ √2x + 1 dx — 用 u = 2x + 1，du = 2 dx ⇒ (1/2) ∫ √u du = (1/2)·(2/3) u3/2 + C = (1/3)(2x + 1)3/2 + C。
(d) ∫₀^(π/2) cos³ x sin x dx — u = -sin x dx. When x = 0, u = 1; x = π/2, u = 0. Integral = -∫₁⁰ u³ du = ∫₀¹ u³ du = 1/4.
(d) ∫₀^(π/2) cos³ x sin x dx — 设 u = cos x，du = -sin x dx。x = 0 时 u = 1；x = π/2 时 u = 0。积分 = -∫₁⁰ u³ du = ∫₀¹ u³ du = 1/4。
Working through these systematically will ensure every variant of substitution in the exercise feels familiar by the time of the assessment.
系统地演练这些题目，能确保在测评时每一种代换变式都变得熟悉自如。
10. Summary and Exam Tips | 总结与考试技巧
Integration by substitution becomes second nature with practice. The key is to identify the inner function g(x) whose derivative sits visibly (or can be easily created) in the integrand. In an exam setting, show all steps clearly: state the substitution, find du, rewrite the integral, and, for definite integrals, change the limits. Even if you cannot evaluate the resulting integral, you will earn method marks for correct substitution and limit changes.
通过练习，换元积分法会变成第二天性。关键在于识别内层函数 g(x)，其导数明显（或可以轻易构造出）出现在被积函数中。在考试环境下，请清晰展示所有步骤：陈述代换、求出 du、重写积分，定积分还需变换上下限。即使最终未能计算出结果，正确的代换与积分限变换也能获得方法分。
Finally, always check your answer by differentiation. This habit, cultivated consistently while doing Exercise 20F.2, will give you confidence and reveal any accidental slip in the integration process.
最后，务必通过求导来检验你的答案。坚持在练习 20F.2 时养成这一习惯，将让你充满信心，并能够及时发现积分过程中的任何意外失误。
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