📚 Mastering Integration by Substitution | 掌握换元积分法
Integration by substitution is one of the most powerful techniques in A-level calculus. It transforms a complicated integral into a simpler one by changing the variable of integration. Whether you are preparing for Edexcel Pure Mathematics Paper 1 or Paper 2, mastering this method is essential for tackling a wide range of functions, from rational expressions to trigonometric and exponential integrands.
换元积分法是 A-level 微积分中最有力的方法之一。它通过改变积分变量，将复杂的积分转化为简单的形式。无论你是在准备 Edexcel 纯数 1 卷还是 2 卷，掌握这一技巧对于处理从有理式到三角、指数被积函数的各类题型都至关重要。
1. Why Substitution Works | 换元法为何有效
The chain rule for differentiation tells us that the derivative of a composite function f(g(x)) is f'(g(x)) · g'(x). Integration by substitution is essentially the reverse of the chain rule, allowing us to "unwind" such derivatives when they appear inside an integral.
微分的链式法则告诉我们，复合函数 f(g(x)) 的导数是 f'(g(x)) · g'(x)。换元积分法本质上是链式法则的逆运算，当积分中出现此类导数结构时，我们便可将其“解开”。
When we set u = g(x), the differential du = g'(x) dx naturally replaces the g' part, leaving a simpler integral in terms of u. This is why the method is sometimes called "u-substitution".
当我们设 u = g(x) 时，微分 du = g'(x) dx 自然地替代了 g'(x) dx 部分，剩下一个关于 u 的更简单积分。因此该方法也常被称为“u 代换法”。
2. Choosing the Right Substitution | 选择合适的代换
The key to success is identifying an "inner function" whose derivative also appears in the integrand, or nearly appears. Look for a function nested inside another, such as a polynomial inside a power, a linear expression inside a trigonometric function, or the exponent of e.
成功的关键在于找出一个“内层函数”，其导数也出现在被积函数中，或近乎出现。寻找嵌套在另一个函数内部的函数，例如幂内的多项式、三角函数内的线性表达式，或者 e 的指数。
t, and u = sin x or cos x when the other trig function and its derivative appear together.
常见的代换包括：线性内层函数用 u = ax + b；当同时存在 x 时用 u = x² + c；当出现另一三角函数及其导数时用 u = sin x 或 cos x。
3. Step-by-Step Procedure for Indefinite Integrals | 不定积分的分步流程
se u and write u = g(x). (2) Compute du/dx and express dx in terms of du, i.e. dx = du / g' ) Integrate with respect to u. (5) Replace u with g(x) to return to the original variable.
对于不定积分，请遵循以下步骤：(1) 选择 u 并写出 u = g(x)。(2) 计算 du/dx，用 du 表示 dx，即 dx = du / g'(x)。(3) 代入所有含 x 的表达式和 dx，得到纯 u 的积分。(4) 对 u 积分。(5) 将 u 替换回 g(x)，返回原变量。
Always remember to include the constant of integration +c at the end of an indefinite integral. Many exam marks are lost for forgetting this simple addition.
务必记住在不打积分末尾加上积分常数 +c。许多考试失分都是因为忘记了这个简单的附加项。
4. Example: ∫ 2x·cos(x² + 1) dx | 示例：∫ 2x·cos(x² + 1) dx
and. The integral becomes ∫ cos u du = sin u + C. Substituting back gives sin(x² + 1) + C.
设 u = x² + 1，则 du/dx = 2x，因此 du = 2x dx。注意到被积函数中正好出现 2x dx。积分变为 ∫ cos u du = sin u + C。回代得到 sin(x² + 1) + C。
This example illustrates the perfect scenario where the derivative of the inner function is sitting beside it. In such cases, substitution is almost automatic.
这个例子展示了一个完美场景：内层函数的导数就紧挨在它旁边。这种情况下，换元几乎是水到渠成。
5. Handling Constants and Adjustments | 处理常数与调整
of u does not appear exactly, but only differs by a constant factor. For example, in ∫ x·√2x²+3 dx. Our integrand contains x dx, so we write x dx = (1/4) du. The integral becomes (1/4) ∫ √u du.
有时 u 的导数并不精确出现，而只是相差一个常数因子。例如，在 ∫ x·√2x²+3 dx 中，令 u = 2x²+3。则 du = 4x dx。我们的被积函数含有 x dx，因此写成 x dx = (1/4) du。积分变为 (1/4) ∫ √u du。
Always check if you can adjust by multiplying and dividing by a constant. This technique is perfectly valid because constants can move in and out of the integral freely.
始终检查是否可以通过乘除常数来调整。这种技巧完全合法，因为常数可以自由地移入或移出积分。
6. Definite Integration by Substitution | 定积分的换元积分
When evaluating a definite integral, you have two options. Option 1: find the indefinite integral in terms of u, change back to x, and then evaluate at the original limits. Option 2 (usually faster): change the limits of integration to match the u-variable.
在计算定积分时，你有两种选择。选择一：求出 u 的不定积分，回代为 x 的表达式，然后代入原上下限求值。选择二（通常更快）：将积分上下限转换为与 u 变量一致的限值。
u = g(a) and u = g(b). Then evaluate the u-integral directly without ever returning to x.
对于选择二，一旦 u = g(x)，计算在下限 (x = a) 和上限 (x = b) 处的 u 值。新限变为 u = g(a) 和 u = g(b)。然后直接计算 u 的积分，无需返回 x。
7. Example: ∫₀¹ 3x²·ex³ dx (Changing Limits) | 示例：∫₀¹ 3x²·ex³ dx（变换上下限）
= 3x² dx. When x = 0, u = 0³ = 0; when x = 1, u = 1³ = 1. The integral transforms to ∫₀¹ eu du = [eu]₀¹ = e¹ – e⁰ = e – 1.
设 u = x³，则 du = 3x² dx。当 x = 0 时，u = 0³ = 0；当 x = 1 时，u = 1³ = 1。积分转换为 ∫₀¹ eu du = [eu]₀¹ = e¹ – e⁰ = e – 1。
Notice how changing the limits avoids the need to substitute back. Always update the limits immediately after deciding on the substitution. Label them clearly to avoid confusion.
请注意，变换上下限后就不需要再回代了。决定代换后要立即更新上下限。清晰地标注出来以免混淆。
8. Recognising "Reverse Chain Rule" Patterns | 识别“反向链式法则”模式
With practice, you will spot integrands of the form k·f'(g(x))·g'(x). These can be integrated directly: ∫ f'(g(x))·g' Substitution is still the formal method, but recognising the pattern speeds up your work.
通过练习，你会识别出形如 k·f'(g(x))·g'(x) 的被积函数。它们可以直接积分：∫ f'(g(x))·g'(x) dx = f(g(x)) + C。代换法仍是规范方法，但识别模式能加快你的解题速度。
For instance, ∫ sin x · ecos x dx. The derivative of cos x is –sin x, so we adjust: –∫ (–sin x)·ecos x dx = –ecos x + C. The inner function is cos x, and its derivative is present.
例如，∫ sin x · ecos x dx。cos x 的导数是 –sin x，因此我们调整：–∫ (–sin x)·ecos x dx = –ecos x + C。内层函数是 cos x，其导数在场。
9. Standard Substitutions for Rational and Radical Functions | 有理函数与根式函数的标准代换
For integrals involving √a² – x², √x² + a², or √x² – a² tions or simpler u = inner part. For rational functions where the numerator is a multiple of the derivative of the denominator, integration gives a natural logarithm: ∫ f'(x)/f(x) dx = ln|f(x)| + C.
对于包含 √a² – x²、√x² + a² 或 √x² – a² 的积分，在进阶数学中常使用三角或双曲代换，但在 A-level (Edexcel) 中主要遇到的是线性代换或简单的内层函数代换。当有理函数的分子是分母导数的倍数时，积分结果为自然对数：∫ f'(x)/f(x) dx = ln|f(x)| + C。
ample, ∫ (2x+1)/(x²+x+5) dx: let u = x²+x+5, du = (2x+1)dx, integral becomes ∫ 1/u du = ln|u| + C.
事实上，对于此类情况，设 u = 分母这一代换非常完美。例如，∫ (2x+1)/(x²+x+5) dx：令 u = x²+x+5，du = (2x+1)dx，积分变为 ∫ 1/u du = ln|u| + C。
10. Substitution with Trigonometric Identities | 结合三角恒等式的代换
ple, ∫ sin³x cos x dx. Let u = sin x, du = cos x dx → ∫ u³ du = u⁴/4 + C = (sin⁴ x)/4 + C.
有时你需要在代换之前或之后使用三角恒等式。例如，∫ sin³x cos x dx。令 u = sin x，du = cos x dx → ∫ u³ du = u⁴/4 + C = (sin⁴ x)/4 + C。
If the powers are both even, you may need to use double-angle formulae to simplify before substituting. Always scan the integrand for potential identities.
如果幂次均为偶数，你可能需要先用倍角公式化简，再进行代换。务必扫描被积函数寻找潜在的恒等式。
11. Common Mistakes and How to Avoid Them | 常见错误及如何避免
Forgetting to express dx in terms of du. Many students write du = g'(x) but leave dx as it is. Always solve for dx explicitly.
忘记将 dx 用 du 表示。许多学生写出 du = g'(x) 却仍保留 dx。务必明确解出 dx。
Not changing limits in definite integrals, or mixing old and new limits. Use one method consistently.
没有更换定积分的上下限，或者混淆旧限与新限。请始终如一地使用一种方法。
Dropping the constant of integration (+c) in indefinite integrals. Always include it.
遗漏不定积分的常数项（+c）。必须始终加上它。
Choosing a u that does not simplify the integral. If the new integral is more complicated, try a different substitution or a different technique.
选择了不能简化积分的 u。如果新的积分更复杂，尝试不同的代换或其他技巧。
12. Exam Tips for Edexcel A-Level Mathematics | Edexcel A-Level 数学考试技巧
Edexcel often embeds substitution within larger problems, such as finding the area under a curve or solving a differential equation. Be prepared to decide on a substitution without being told. Exercises from the textbook and past papers (e.g., 9MA0/01) will hone this skill.
Edexcel 常在较大的题目中嵌入代换，例如求曲线下面积或解微分方程。准备好自己去决定代换，而不是等提示。教材和历年真题（如 9MA0/01）的练习将磨练此项技能。
Always show your substitution and the transformation of the differential clearly. Even if the final answer is wrong, clear method marks can be awarded for choosing u, finding du, and converting the integral correctly.
务必清晰地写出代换过程以及微分的变换。即使最终答案错误，只要明确选择了 u、求出 du 并正确转换了积分，就能获得方法分。
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