Exercise 20B.2: Integration by Substitution | 练习 20B.2：代换积分法

📚 Exercise 20B.2: Integration by Substitution | 练习 20B.2：代换积分法

Integration by substitution is one of the most powerful techniques in the IB Mathematics Analysis & Approaches syllabus. In Exercise 20B.2, you are challenged to apply the reverse chain rule to a variety of functions, transforming complex integrals into standard forms. This article provides a comprehensive walkthrough of the method, common pitfalls, and step-by-step examples that mirror the style of questions you will encounter in your problem set.

代换积分法是 IB 数学分析与方法课程中最强有力的技巧之一。在练习 20B.2 中，你将面临各种函数的挑战，需要运用反向链式法则，将复杂的积分转化为标准形式。本文全面讲解该方法、常见误区，并提供分步范例，与你习题集中会遇到的问题风格高度一致。

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1. The Core Idea of Substitution | 代换法的核心思想

pression that matches one of the standard integrals from the IB formula booklet, such as ∫ uⁿ du or ∫(1/u) du.

代换法，又称 u 换元法，能够帮助我们抵消微分的链式法则。如果一个被积函数包含复合函数，且其内层函数的导数作为因子出现，我们就可以设 u = g(x)，并将积分用 u 表示。目标是得到与 IB 公式手册中标准积分相符的表达式，例如 ∫ uⁿ du 或 ∫(1/u) du。

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2. Recognising When to Use Substitution | 识别何时使用代换法

Look for integrals where the integrand can be expressed as f(g(x)) · g'(x). Common signs are a linear expression raised to a power multiplied by its derivative, trigonometric functions with a linear argument, or a fraction whose numerator resembles the derivative of its denominator. In Exercise 20B.2, problems often feature √ax + b, sin(ax + b), or e^{ax + b} combined with a constant factor that can be adjusted to match g'(x).

寻找被积函数可以表示为 f(g(x)) · g'(x) 的积分。常见的迹象有：某个线性表达式的幂乘以其导数、自变量为线性的三角函数，或者分子像是分母的导数的分式。在练习 20B.2 中，题目常常包含 √ax + b、sin(ax + b) 或 e^{ax + b} 与一个常数因子的组合，该常数可调整以匹配 g'(x)。

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3. The Systematic Procedure | 系统化步骤

Choose u = g(x), usually the inner part of a composite function. Step 2: Compute du/dx = g'(x) and rearrange to obtain dx = du / g'(x). Step 3: Substitute both u and dx into the integral, cancelling the g' with g(x) to return to the original variable. Always add the constant of integration + C.

遵循清晰的四步流程。第一步：选择 u = g(x)，通常是复合函数的内层部分。第二步：计算 du/dx = g'(x) 并整理得到 dx = du / g'(x)。第三步：将 u 与 dx 一同代入积分，若选择正确则可消去 g'(x) 项。第四步：对 u 积分，然后用 g(x) 替换 u 回到原变量。切勿忘记添加积分常数 + C。

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4. Worked Example 1: Power of a Linear Function | 范例一：线性函数的幂

/ 15 + C. This demonstrates how a missing constant factor is handled by adjusting the differential.

考虑 ∫ (3x + 5)⁴ dx。这是一道典型的练习 20B.2 题目。设 u = 3x + 5，则 du/dx = 3，dx = du/3。代入得：∫ u⁴ (du/3) = (1/3) ∫ u⁴ du = (1/3) · (u⁵/5) + C = u⁵/15 + C。回代 u： (3x + 5)⁵ / 15 + C。此例展示了如何通过调整微分来处理缺失的常数因子。

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5. Worked Example 2: Trigonometric Function | 范例二：三角函数

ch does not appear explicitly, but we can introduce the factor 4 by writing dx = (1/4) du.

积分 ∫ cos(4x) dx。设 u = 4x，得 du/dx = 4，dx = du/4。积分变为 ∫ cos(u) (du/4) = (1/4) sin(u) + C。回代后得 (1/4) sin(4x) + C。注意内层函数 4x 的导数 4 并未显式出现，但我们可以通过将 dx 写成 (1/4) du 来引入因子 4。

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6. Adjusting Constants: The Missing Factor | 调整常数：缺失的因子

estions, the derivative of the inner function is not perfectly present. For instance, ∫ x² √x³ + 1 dx. Let u = x³ + 1, then du/dx = 3x², so x² dx = du/3. Substitute: ∫ √u (du/3) = (1/3) ∫ u^1/2 du = (1/3) · (2/3) u^3/2 + C = (2/9)(x³ + 1)^3/2 + C. The variable part x² dx is replaced by du/3, perfectly matching the substitution.

在许多练习 20B.2 的题目中，内层函数的导数并不完美地出现。例如，∫ x² √x³ + 1 dx。设 u = x³ + 1，则 du/dx = 3x²，所以 x² dx = du/3。代入得：∫ √u (du/3) = (1/3) ∫ u^1/2 du = (1/3) · (2/3) u^3/2 + C = (2/9)(x³ + 1)^3/2 + C。变量部分 x² dx 被 du/3 替代，与代换完美匹配。

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7. Definite Integrals and Changing Limits | 定积分与积分限的变换

efore applying the original limits. The first method is cleaner. For ∫ from 0 to 1 of 2x e^x² du = 2x dx. When x = 0, u = 0; when x = 1, u = 1. The integral becomes ∫ from 0 to 1 of e^u du = [e^u] from 0 to 1 = e − 1. There is no need to return to x.

使用代换法解定积分时有两个选择：要么在计算前将积分限变换为 u 变量，要么先在 u 下积分后回代为 x 再代入原积分限。第一种方法更为简洁。对于 ∫₀¹ 2x e^x² dx，设 u = x²，du = 2x dx。当 x = 0 时 u = 0；x = 1 时 u = 1。积分成为 ∫₀¹ e^u du = [e^u]₀¹ = e − 1，无需回到 x。

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8. Recognising Reverse Substitutions | 识别反向代换

Sometimes the substitution is less obvious, such as when the integrand contains √a² − x². In IB you might be guided to use x = a sin θ. For instance, ∫ √4 − x² dx can be tackled with x = 2 sin θ, dx = 2 cos θ dθ. Then √4 − x² = √4 − 4 sin²θ θ dθ, which can be solved using double-angle identities. This appears in later sections, but the principle of choosing a substitution that simplifies the root is vital.

有时代换不那么明显，例如被积函数包含 √a² − x² 的情况。IB 中可能会引导你使用 x = a sin θ。比如，∫ √4 − x² dx 可通过设 x = 2 sin θ, dx = 2 cos θ dθ 来处理。那么 √4 − x² = √4 − 4 sin²θ = 2 cos θ。积分变为 ∫ 4 cos²θ dθ，可用二倍角公式求解。这出现在后续章节，但选择能简化根式的代换这一原则至关重要。

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9. Common Mistakes in Exercise 20B.2 | 练习 20B.2 中的常见错误

dx in terms of du. Another mistake is mishandling the constant factor: if du contains a coefficient, the reciprocal must multiply the integral. Also, after integrating, some forget to substitute back to x. In definite integrals, failing to update the limits is a serious error. Double-check each step by differentiating your answer; it should yield the original integrand.

一个常见错误是忘记完全替换 dx。学生常写成 ∫ f(u) dx，忽略了用 du 表示 dx。另一个错误是常数因子处理不当：如果 du 中包含系数，其倒数须乘以整个积分。此外，积分后有人忘记回代为 x。在定积分中，未更新积分限是严重的失误。通过对答案求导来逐步检查，结果应等于原被积函数。

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10. Substitution with Exponential Functions | 指数函数的代换

Exponential integrals frequently appear in Exercise 20B.2. For ∫ e^2x+1 dx, let u = 2x+1, du = 2 dx, giving (1/2) e^u + C = (1/2) e^2x+1 + C. More complex cases like ∫ x e^x² dx use u = x², du = 2x dx, transforming to (1/2) ∫ e^u du. Always look for an exponent whose derivative is present as a factor. If the derivative is missing but can be created by inserting a constant, the substitution is valid after adjusting.

指数函数的积分在练习 20B.2 中频繁出现。对于 ∫ e^2x+1 dx，设 u = 2x+1，du = 2 dx，得到 (1/2) e^u + C = (1/2) e^2x+1 + C。更复杂的情况如 ∫ x e^x² dx 使用 u = x²，du = 2x dx，转化为 (1/2) ∫ e^u du。始终寻找其指数函数的导数作为因子出现的情况。若导数缺失但可通过插常数生成，经调整后代换依然有效。

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11. Using Substitution to Derive Standard Forms | 利用代换推导标准形式

Substitution helps prove results like ∫ (f'(x)/f(x)) dx = ln |f(x)| er ∫ (2x/(x² + 1)) dx. Set u = x² + 1, du = 2x dx, so the integral becomes ∫ (1/u) du = ln |u| re the numerator is the derivative of the denominator. Recognising it saves time in exams.

代换法有助于证明诸如 ∫ (f'(x)/f(x)) dx = ln |f(x)| + C 等结果。在你的习题集中，可能会遇到 ∫ (2x/(x² + 1)) dx。设 u = x² + 1，du = 2x dx，积分变为 ∫ (1/u) du = ln |u| + C = ln(x² + 1) + C。此模式在分子为分母导数的有理函数中经常出现，认出它能在考试中节省时间。

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12. Practice Strategies and IB Exam Tips | 练习策略与 IB 考试技巧

Work through all parts of Exercise 20B.2 methodically, starting with the simpler linear substitutions before tackling those requiring algebraic manipulation. Always verify by differentiation. In IB exams, questions often provide a substitution instruction, but you must be able to select your own when it is not given. Create a summary table of common substitutions: linear expressions, powers, exponentials, and trigonometric functions. Regular practice builds the intuition needed to see the inner function instantly.

有条不紊地完成练习 20B.2 的所有部分，从较简单的线性代换开始，再处理需要代数变形的题目。始终通过求导加以验证。在 IB 考试中，题目常会给出代换指示，但未提供时你必须能自己选择代换。制作一张常见代换的汇总表：线性表达式、幂函数、指数函数和三角函数。经常练习能培养瞬间看出内层函数的直觉。

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