📚 Integration by Substitution | 换元积分法
Integration by substitution is one of the most powerful techniques in A‑Level Edexcel Pure Mathematics. It allows us to transform a complicated integral into a standard form by changing the variable of integration. Mastering this method is essential for tackling a wide range of integrals that appear in Paper 1 and Paper 2, from polynomials to trigonometric, exponential, and rational functions.
换元积分法是 A‑Level Edexcel 纯数学中最强有力的积分技巧之一。它通过更换积分变量，将复杂的被积函数转化为标准形式。掌握这一方法对于应对卷一卷二中出现的多项式、三角函数、指数函数和有理函数的积分至关重要。
1. The Idea Behind Substitution | 换元法的核心思想
Substitution is the reverse process of the chain rule for differentiation. When we differentiate a composite function y = f(g(x)), the derivative is dy/dx = f'(g(x)) · g'(x). Integration by substitution systematically undoes this by letting u = g(x), so that du/dx = g'(x) and du = g'(x) dx. The goal is to rewrite the entire integral in terms of u, making it easier to integrate.
换元法是链式法则的逆过程。对复合函数 y = f(g(x)) 求导得 dy/dx = f'(g(x)) · g'(x)。换元积分法通过设 u = g(x) 系统地逆转这一过程，使得 du/dx = g'(x)，从而 du = g'(x) dx。目标是整个积分用 u 表示，使积分变得容易计算。
2. The Standard Procedure | 标准操作步骤
The method can be broken down into four clear steps. First, choose a suitable substitution u = g(x), often the inner function of a composite. Second, compute du/dx and rearrange to replace dx with du. Third, rewrite the entire integrand and the differential dx in terms of u — there should be no x’s left. Fourth, integrate with respect to u, and finally substitute back to the original variable.
该方法可拆分为四个清晰的步骤。首先选择合适的代换 u = g(x)，通常选取复合函数的内层部分。其次计算 du/dx 并将 dx 替换为 du。第三步将被积函数和微分 dx 完全用 u 表示——最后不应残留任何 x。最后关于 u 积分，再将变量代回原变量。
3. Example 1 – A Simple Polynomial | 例1 – 简单多项式
Consider the integral ∫ 2x (x² + 1)⁴ dx. Let u = x² + 1. Then du/dx = 2x, so du = 2x dx bstituting transforms the integral into ∫ u⁴ du. Integrating gives (1/5) u⁵ + C. Replacing u, the final answer is (1/5)(x² + 1)⁵ + C.
考虑积分 ∫ 2x (x² + 1)⁴ dx。令 u = x² + 1，则 du/dx = 2x，即 du = 2x dx。代换后将积分变为 ∫ u⁴ du。积分得 (1/5) u⁵ + C。将 u 代回，最终结果为 (1/5)(x² + 1)⁵ + C。
4. Example 2 – Trigonometric Substitution | 例2 – 三角代换
ate ∫ sin³x cos x dx. Notice that the derivative of sin x is cos x, so a natural choice is u = sin x. Then du = cos x dx s how substitution turns a product of a power of a trig function and its derivative into a simple power integral.
计算 ∫ sin³x cos x dx。注意到 sin x 的导数是 cos x，因此自然选择 u = sin x。于是 du = cos x dx。积分变为 ∫ u³ du = (1/4) u⁴ + C = (1/4) sin⁴x + C。这呈现了代换如何将三角函数的幂与其导数之积转化为简单的幂积分。
5. Example 3 – Exponential Functions | 例3 – 指数函数
For ∫ e2x+3 dx, let u = 2x + 3. Hence du/dx = 2, giving dx = du/2. The integral becomes ∫ eu (du/2) = (1/2) eu + C = (1/2) e2x+3 + C. This pattern is extremely common in Edexcel exam questions involving exponential growth and decay.
对于 ∫ e2x+3 dx，令 u = 2x + 3。因此 du/dx = 2，得 dx = du/2。积分变为 ∫ eu (du/2) = (1/2) eu + C = (1/2) e2x+3 + C。这种模式在 Edexcel 考试中涉及指数增长与衰减的问题中极为常见。
6. Example 4 – A Rational Function | 例4 – 有理函数
Find ∫ x / √x – 1 dx. Set u = x – 1, so du = dx and x = u + 1. The integral becomes ∫ (u + 1) / √u du = ∫ (u1/2 + u–1/2) du. Integrating term by term gives (2/3) u3/2 + 2 u1/2 + C = (2/3)(x – 1)3/2 + 2(x – 1)1/2 + C. This demonstrates how substitution can handle expressions with square roots by rationalising the integrand.
求 ∫ x / √x – 1 dx。设 u = x – 1，则 du = dx 且 x = u + 1。积分变为 ∫ (u + 1) / √u du = ∫ (u1/2 + u–1/2) du。逐项积分得 (2/3) u3/2 + 2 u1/2 + C = (2/3)(x – 1)3/2 + 2(x – 1)1/2 + C。这一例子展示了代换如何处理含根号的表达式，将被积函数有理化。
7. Substitution for Definite Integrals | 定积分的换元法
When handling a definite integral, there are two approaches. You can either change the limits to match the substitution, or integrate in terms of u and then substitute back the original limits. With the first method, once you let u = g(x), compute the new limits u₁ = g(a) and u₂ = g(b) and integrate entirely in u without returning to x. For example, ∫₀¹ 2x(x²+1)⁴ dx with u = x²+1 gives limits u = 1 to u = 2, then ∫₁² u⁴ du = [u⁵/5]₁² = 32/5 – 1/5 = 31/5.
处理定积分时有两种方法。既可以相应更换积分上下限，也可以先关于 u 积分最后代回原上下限。第一种方法：设 u = g(x) 后，计算出新限值 u₁ = g(a) 与 u₂ = g(b)，全程在 u 下积分而不回到 x。例如 ∫₀¹ 2x(x²+1)⁴ dx，令 u = x²+1 得到上下限 u = 1 到 u = 2，于是 ∫₁² u⁴ du = [u⁵/5]₁² = 32/5 – 1/5 = 31/5。
8. Choosing the Right Substitution | 正确选取代换
Selecting the most effective substitution comes with practice. Generally, look for a function whose derivative also appears in the integrand. If you see a composite function, try setting u to be the inner function. For trigonometric integrals, common choices are u = sin x, u = cos x, or u = tan x. For exponentials, the exponent itself is a good candidate. If a square root contains a linear expression, substituting the whole linear expression often simplifies the integral.
选取最有效的代换需要通过练习培养。一般而言，寻找其导数也出现在被积函数中的函数。若看到复合函数，尝试将内层函数设为 u。对于三角积分，常见选择是 u = sin x、u = cos x 或 u = tan x。对于指数函数，指数本身是一个好选择。若平方根内是线性表达式，替换整个线性表达式往往能简化积分。
9. Common Pitfalls and How to Avoid Them | 常见陷阱与规避方法
A frequent mistake is forgetting to replace dx entirely with du. Always solve for dx in terms of du and check that no x remains after substitution. Another error is misapplying the chain rule when differentiating the substitution; be meticulous when working out du/dx. Also, when dealing with definite integrals, failing to change the limits can lead to incorrect answers. Finally, always include the constant of integration + C for indefinite integrals — marks are often lost here.
常见错误包括忘记将 dx 完全替换为 du。务必解出 dx 关于 du 的表达式，并检查代换后是否已无 x。另一个错误是在对代换式求导时误用链式法则；计算 du/dx 时应格外仔细。处理定积分时未更换上下限也会导致错误答案。最后，不定积分务必加上积分常数 + C——这里常常失分。
10. Recognising When Not to Use Substitution | 识别何时不需要换元
x² + 2x) dx is trivial by the power rule. Similarly, ∫ (sin x)/cos x dx is a standard log integral once you spot the derivative relationship. Substitution is a tool, not a universal strategy — use it where it genuinely simplifies the integral.
并非每个积分都需要换元。有些通过标准模式或代数处理就能更高效地求解。例如 ∫ (3x² + 2x) dx 只需运用幂规则。类似地，一旦注意到导数关系，∫ (sin x)/cos x dx 就是一个标准对数积分。换元是一种工具而非万能策略——应在其真正简化积分时使用。
11. Practice Skills and Exam Tips | 练习技巧与应试建议
When revising, work through integrals that mix polynomials, trig functions, exponentials, and rational functions. Edexcel frequently combines substitution with other techniques such as integration by parts or partial fractions. In exams, show your substitution clearly, write the transformed integral, and remember to change limits if solving a definite integral. Practising under timed conditions will build speed and confidence.
复习时要练习混合多项式、三角函数、指数函数和有理函数的积分。Edexcel 常将代换与分部积分或部分分式等其他技巧结合考查。考试中应清晰写出代换过程、变形后的积分，若求定积分则记住更换上下限。在限时条件下练习可提升速度和信心。
12. Summary Table of Common Substitutions | 常见代换总结表
The table below summarises frequently used substitutions for different integrand forms. Familiarity with these will help you quickly identify the right approach in an exam.
下表汇总了不同被积函数形式下常用的代换。熟悉这些将有助于你在考试中迅速识别正确方法。
| Integrand Form | Suggested Substitution | Reason |
|---|---|---|
| f(ax+b) | u = ax + b | du = a dx, simple linear shift |
| f'(x) · [f(x)]ⁿ | u = f(x) | Derivative matches the factor |
| sinⁿx cos x | u = sin x | cos x dx = du |
| cosⁿx sin x | u = cos x | –sin x dx = du (manage sign) |
| ekx | u = kx | dx = du/k |
| 1/(x ln x) | u = ln x | du = (1/x) dx |
| √ax+b in denominator | u = √ax+b or u = ax+b | Simplifies root expression |
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