Integration by Substitution | 积分换元法

📚 Integration by Substitution | 积分换元法

Integration by substitution is one of the most versatile techniques in A-Level calculus. It transforms a complicated integral into a simpler one by changing the variable, much like reversing the chain rule. Mastering substitution not only unlocks a wide range of exam questions but also deepens your understanding of how differentiation and integration are connected.

积分换元法是 A-Level 微积分中最灵活的技巧之一。它通过变量替换，将复杂的积分转化为简单的形式，本质上就是链式求导法则的逆运算。掌握换元法不仅能解锁大量考试题型，还能加深你对微分与积分内在联系的理解。

1. Introduction to Integration by Substitution | 积分换元法简介

When we differentiate a composite function y = f(g(x)) we use the chain rule: dy/dx = f'(g(x)) · g'(x). Therefore, to integrate an expression that looks like f'(g(x)) · g' rocess. Substitution provides a systematic way to do this: we set u = g(x), then du/dx = g'(x), so du = g'(x) dx. The integral ∫ f'(g(x)) · g'(x) dx becomes ∫ f'(u) du, which is straightforward.

当我们对复合函数 y = f(g(x)) 求导时，使用链式法则：dy/dx = f'(g(x)) · g'(x)。因此，要积分形如 f'(g(x)) · g'(x) 的表达式，可以逆用这一过程。换元法提供了一个系统的方法：令 u = g(x)，则 du/dx = g'(x)，于是 du = g'(x) dx。原积分 ∫ f'(g(x)) · g'(x) dx 变为 ∫ f'(u) du，这就容易计算了。

y be given the substitution, or you may have to spot it yourself. Typical instructions are 'Using the substitution u = ...' or 'Use the substitution ... to find' oth indefinite and definite integrals, with careful handling of limits in the latter case.

在 Edexcel 试卷中，题目可能会给出换元式子，也可能需要你自己发现。常见指令如“利用换元 u = ... 求积分”。该技术对不定积分和定积分均适用，但在定积分中需要谨慎处理上下限。

2. The Reverse Chain Rule | 逆链式法则

s useful to recognise the reverse chain rule directly. If an integral has the form ∫ k · f'(g(x)) · g' the answer as k · f(g(x)) + c. This is often tested in Edexcel AS and A-Level exams under 'integration by inspection'.

在正式学习换元法之前，先直接识别逆链式法则很有帮助。如果积分具有 ∫ k · f'(g(x)) · g'(x) dx 的形式，我们可以直接写出答案为 k · f(g(x)) + c。这在 Edexcel AS 和 A-Level 考试中常以“观察法积分”出现。

e shows that whenever the derivative of the inner function sits as a factor, substitution is extremely quick.

例如，考虑 ∫ 2x · cos(x²) dx。令 u = x²，则 du = 2x dx，积分变为 ∫ cos u du = sin u + c = sin(x²) + c。逆链式法则表明，当内部函数的导数作为因子出现时，换元变得极其迅速。

∫ f'(g(x)) · g'(x) dx = f(g(x)) + c

3. Choosing the Substitution | 选择换元

we decide on a substitution u = g(x). The goal is to simplify the integrand so that all x's and dx are expressed in terms of u and du. Usually we set u to be an 'inner' function that makes the integral less messy – ten the argument of a bracket, a trigonometric function, an exponential, or a square root.

当积分不是直接的逆链式法则时，我们需要选择换元 u = g(x)。目标是简化被积函数，使所有的 x 和 dx 都用 u 和 du 表示。通常我们把 u 设为能使积分变简洁的“内部”函数——常见的有括号内的表达式、三角函数或指数函数的自变量、平方根内部等。

For instance, if we see √3x+1, we might set u = 3x+1. If we have e^5x 2x-3)⁶ appears multiplied by something close to its derivative, the substitution is clear. Sometimes Edexcel questions provide the substitution; you then need to find du/dx, rearrange to get dx in terms of du, and replace all x-parts.

比如，若见到 √3x+1，我们可以设 u = 3x+1。如果出现 e^5x，可设 u = 5x。若是 (2x-3)⁶ 且乘积接近其导数，换元便十分明显。有时 Edexcel 题目提供换元式；你则需计算 du/dx，重新整理得到 dx 关于 du 的表达式，并替换所有含 x 的部分。

4. Substitution with Polynomials | 多项式的换元

tituting gives ∫ (u³)·(½ du) = ½ ∫ u³ du = ½ · u⁴/4 + c = (1/8) u⁴ + c = (1/8)(x²+4)⁴ + c.

考虑不定积分 ∫ x·(x²+4)³ dx。令 u = x²+4，则 du/dx = 2x，于是 x dx = ½ du。代入后得到 ∫ (u³)·(½ du) = ½ ∫ u³ du = ½ · u⁴/4 + c = (1/8) u⁴ + c = (1/8)(x²+4)⁴ + c。

Notice how the x factor was absorbed by du. This is a key indicator that the substitution will work: the remaining x's cancel out. Always check that after substitution, no x terms remain; otherwise the substitution might not be appropriate unless further manipulation can be done.

注意多余的 x 因子如何被 du 吸收。这是换元可行的重要标志：剩余的 x 项消去。换元后务必检查是否还有 x 变量残留；否则该换元可能不适用，除非能进一步处理。

For definite integrals with polynomial substitutions, the same steps apply, but limits must be converted. We will cover this shortly.

对于使用多项式换元的定积分，步骤相同，但积分上下限必须转换。稍后会讲解。

5. Trigonometric Substitutions | 三角换元

in x dx, or ∫ tan x dx. In ∫ sin³x cos x dx, letting u = sin x gives du = cos x dx, turning the integral into ∫ u³ du.

Edexcel 常考查使用换元简化三角被积函数的情况。常见例子有 ∫ sin(ax+b) dx、∫ cos²x sin x dx 或 ∫ tan x dx。在 ∫ sin³x cos x dx 中，令 u = sin x，则 du = cos x dx，积分变成 ∫ u³ du。

ke sin²x + cos²x = 1) that allow you to rewrite the integral entirely in terms of u. For example, to integrate sin³x, rewrite sin³x = sin x · sin²x = sin x (1 - cos²x), then use u = cos x.

有时题目会给出换元，如对于含 sin²x cos x 的积分设 u = cos x。留意三角恒等式（如 sin²x + cos²x = 1），它们可以将被积函数完全用 u 表示。例如，要积分 sin³x，可改写 sin³x = sin x · sin²x = sin x (1 - cos²x)，然后设 u = cos x。

∫ tan x dx = ∫ (sin x / cos x) dx → let u = cos x → -∫ du/u = -ln|u| + c = -ln|cos x| + c

6. Definite Integrals and Changing Limits | 定积分与换限

When evaluating a definite integral via substitution, you have two options: change the limits to the new variable u and never return to x, or compute the indefinite integral in terms of u, substitute back to x, and then apply the original limits. Edexcel mark schemes favour changing limits because it is faster and reduces errors.

在用换元法计算定积分时，你有两种选择：将积分上下限转换为新变量 u 并全程不回到 x；或者求出关于 u 的不定积分后代回 x，再代入原限。Edexcel 的评分方案更倾向于换限，因为它更快且减少错误。

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To change limits: if u = g(x), the lower limit u_lower = g(a) and upper limit u_upper = g(b). After substitution, the integral becomes ∫_ulower^u_upper f(u) du. Then evaluate directly. For example, evaluate ∫₀² x·e^x² dx. Let u = x², du = 2x dx. Limits: when x=0, u=0; when x=2, u=4. The integral becomes ½ ∫₀⁴ e^u du = ½ [e^u]₀⁴ = ½(e⁴ - 1).

换限方法：若 u = g(x)，则下限 u_lower = g(a)，上限 u_upper = g(b)。经换元后，积分变为 ∫_ulower^u_upper f(u) du，然后直接计算。例如，求 ∫₀² x·e^x² dx。令 u = x²，则 du = 2x dx。上下限：x=0 时 u=0；x=2 时 u=4。原积分 = ½ ∫₀⁴ e^u du = ½ [e^u]₀⁴ = ½(e⁴ - 1)。

Always write the new limits clearly and never mix x and u in the same integral after substitution.

换元后始终要清晰写出新限，不要在积分中混用 x 和 u。

7. Exponential and Logarithmic Substitutions | 指数与对数换元

Exponential functions are straightforward candidates for substitution. For ∫ e^ax+b dx, use u = ax+b, then du = a dx. A more interesting case is ∫ x·e^2x titution alone is insufficient. However, Edexcel may ask you to use the substitution u = e^x in integrals involving e^2x or rational functions of e^x.

指数函数是换元的直接候选对象。对于 ∫ e^ax+b dx，可设 u = ax+b，则 du = a dx。更有意思的是 ∫ x·e^2x dx，但它需要分部积分；仅靠换元不够。不过，Edexcel 可能会要求对包含 e^2x 或 e^x 的有理函数的积分使用换元 u = e^x。

For example, ∫ e^x / √1+e^2x dx. Let u = e^x, then du = e^x dx, so the integral becomes ∫ 1/√1+u² du. You can then recognise a standard inverse hyperbolic or trigonometric substitution (depending on the module).

例如，∫ e^x / √1+e^2x dx。令 u = e^x，则 du = e^x dx，积分变为 ∫ 1/√1+u² du。接着可以识别为标准反双曲或三角换元（取决于学习模块）。

/x dx are perfect for u = ln x. Then du = (1/x) dx, transforming the integral into ∫ u du.

对数积分如 ∫ (ln x)/x dx 非常适合设 u = ln x。则 du = (1/x) dx，积分变为 ∫ u du。

8. More Challenging Substitutions | 更复杂的换元

In A2 Edexcel pure maths, substitution can become more involved. A typical question provides the substitution and you must manipulate the integrand using algebraic identities. For instance, to integrate 1/√a²-x² dx, use x = a sin θ, then dx = a cos θ dθ, and √a²-x² = a cos θ. This yields ∫ a cos θ / (a cos θ) dθ = ∫ dθ = θ + c = arcsin(x/a) + c.

在 A2 Edexcel 纯数学中，换元会变得更复杂。典型题型会给出换元式，你需要利用代数恒等变换被积函数。例如，积分 1/√a²-x² dx，设 x = a sin θ，则 dx = a cos θ dθ，且 √a²-x² = a cos θ。于是积分变为 ∫ a cos θ / (a cos θ) dθ = ∫ dθ = θ + c = arcsin(x/a) + c。

Always check the domain and sign of square roots when using trigonometric substitutions. Edexcel usually specifies the range of θ to ensure positive roots. Similarly, hyperbolic substitutions may appear in Further Maths, but are not required in standard A-Level Mathematics.

使用三角换元时要注意定义域和平方根的正负号。Edexcel 通常会指定 θ 的范围以保正根。同样，双曲换元可能出现在进阶数学中，但标准 A-Level 数学不要求。

e the substitution to find x in terms of u, or to express dx in terms of du and u. For u = √x+1, you would square both sides, isolate x, and differentiate. These steps need careful handling.

考试中有一种棘手情况：你需要根据换元重新整理出 x 关于 u 的表达式，或将 dx 用 du 和 u 表示。若 u = √x+1，需两边平方，解出 x，再求导。这些步骤必须细心处理。

9. Common Mistakes to Avoid | 常见错误

rite dx = ... du and replace it. Leaving dx in the integral is a frequent source of error.

(1) 忘记将 dx 换成 du：求出 du/dx 后，必须写出 dx = ... du 并替换。在积分中留下 dx 是常见的错误来源。

(2) Not changing limits in definite integrals: mixing u and x leads to nonsense. Always convert limits when using a u-substitution on a definite integral.

(2) 在定积分中未换限：混用 u 和 x 会导致错误。对定积分使用 u 换元时，务必转换上下限。

(3) Substitution not eliminating all x's: after replacement, if x still appears, double-check the algebra. The substitution should eliminate the original variable.

(3) 换元后未消去所有 x：代换后若 x 仍然出现，请复查代数运算。换元应当消去原变量。

(4) Ignoring the constant of integration in indefinite integrals. Always add +c at the final step.

(4) 不定积分漏掉积分常数。务必在最后一步加上 +c。

(5) Incorrect handling of coefficient when du = k dx. Many students forget to divide by k.

(5) 错误处理 du = k dx 的系数。许多学生忘记除以 k。

Reviewing these common pitfalls will secure marks that are often lost unnecessarily.

复习这些常见陷阱，可保住许多常因粗心丢失的分数。

10. Exam Tips and Summary | 考试提示与总结

In Edexcel exams, substitution questions usually carry 5–8 marks. Show every step methodically: state the substitution, find du/dx, rearrange for dx, change limits if necessary, convert integrand, integrate, and substitute back or evaluate with new limits. Even if your final answer is wrong, clear working can earn method marks.

在 Edexcel 考试中，换元题通常占 5–8 分。要有条理地展示每一步：写出换元式，求 du/dx，整理出 dx，必要时换限，转化被积函数，积分，然后回代或用新限求值。即使最终答案错误，清晰的步骤也可获得方法分。

Memorise standard integrals such as ∫ uⁿ du, ∫ 1/u du, ∫ e^u ing blocks after substitution. Also practise recognising when a given substitution simplifies the expression; sometimes the hardest part is knowing when not to use substitution.

熟记标准积分，如 ∫ uⁿ du, ∫ 1/u du, ∫ e^u du, ∫ cos u du。这些是换元后的基本模块。还要练习识别何时给定的换元能使表达式简化；有时最难的在于知道何时不该用换元。

Common Substitution	Useful for
u = ax + b	Linear inner function
u = x² + constant	Derivative x present
u = sin x or cos x	Powers of trig functions
u = eˣ	Rational functions of eˣ
x = a sin θ	√a² - x²

Ultimately, integration by substitution is about pattern recognition, algebraic fluency, and meticulous execution. With consistent practice, it becomes a reliable tool in your mathematical toolkit.

归根结底，积分换元法在于模式识别、代数流畅和严谨的执行。通过持续练习，它将成为你数学工具箱中一个可靠的工具。

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