📚 A-Level Edexcel Maths: Mastering Integration by Substitution | A-Level Edexcel 数学:掌握换元积分法
Integration by substitution is one of the most powerful techniques in A-Level Edexcel Pure Mathematics. It allows you to transform complicated integrals into simpler standard forms by changing the variable. Mastering this method is essential for both Paper 1 and Paper 2 questions involving integration, trigonometric functions, exponentials, and rational functions.
换元积分法是 A-Level Edexcel 纯数学中最强大的技巧之一。它通过更换变量,把复杂的积分转化为更简单的标准形式。掌握这一方法对于 Paper 1 和 Paper 2 中涉及积分、三角函数、指数函数和有理函数的问题至关重要。
1. What is Integration by Substitution? | 什么是换元积分法?
Integration by substitution is the reverse process of the chain rule in differentiation. When an integrand contains a function and its derivative, we can replace the inner function with a new variable u, rewrite the integral in terms of u, integrate, and then substitute back to the original variable. The goal is to reduce a difficult integral to one of the standard forms such as ∫ uⁿ du, ∫ eᵘ du, or ∫ cos u du.
换元积分法是微分中链式法则的逆过程。当被积函数包含一个函数及其导数时,我们可以把内层函数替换为一个新变量 u,用 u 重写积分,先积分,再代回原变量。这样做的目的是把困难的积分化简为标准形式,例如 ∫ uⁿ du、∫ eᵘ du 或 ∫ cos u du。
∫ f(g(x)) g'(x) dx = ∫ f(u) du, where u = g(x)
∫ f(g(x)) g'(x) dx = ∫ f(u) du,其中 u = g(x)
2. Why Substitution Works: Reverse Chain Rule | 为什么换元法有效:反向链式法则
The chain rule states that d/dx [F(g(x))] = F'(g(x)) g'(x). Therefore, integrating both sides gives ∫ F'(g(x)) g'(x) dx = F(g(x)) + C. When we set u = g(x), we have du/dx = g'(x), which can be written as du = g'(x) dx. This transforms the left-hand side into ∫ F'(u) du = F(u) + C.
链式法则指出 d/dx [F(g(x))] = F'(g(x)) g'(x)。因此,对两边积分可得 ∫ F'(g(x)) g'(x) dx = F(g(x)) + C。当我们令 u = g(x) 时,有 du/dx = g'(x),也可写成 du = g'(x) dx。这样左边就转化为 ∫ F'(u) du = F(u) + C。
This relationship shows that substitution is not a trick, but a direct consequence of reversing differentiation. In exam solutions, you should always state your substitution u, find du/dx, and show the replacement of dx before integrating.
这一关系说明换元法不是一种技巧,而是逆向微分的直接结果。在考试解答中,你应当始终写出换元变量 u,求出 du/dx,并在积分前展示 dx 的替换过程。
3. Choosing the Substitution u | 如何选择换元变量 u
Look for an inner function whose derivative also appears in the integrand. A good substitution usually cancels out the remaining x-term or simplifies the expression. Common choices include u = x² + 1 when x appears alongside, u = sin x when cos x is present, and u = ln x when 1/x appears.
寻找一个内层函数,并且它的导数也出现在被积函数中。一个好的换元通常会约掉剩余的 x 项或简化表达式。常见的选择包括:当旁边有 x 时令 u = x² + 1;当有 cos x 时令 u = sin x;当有 1/x 时令 u = ln x。
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If the integrand contains x and x² + a², try u = x² + a², so du/dx = 2x.
如果被积函数含有 x 和 x² + a²,尝试令 u = x² + a²,因此 du/dx = 2x。
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If the integrand contains cos x and a power of sin x, try u = sin x, so du/dx = cos x.
如果被积函数含有 cos x 和 sin x 的幂,尝试令 u = sin x,因此 du/dx = cos x。
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If the integrand contains eˣ or 1/x and a function of eˣ or ln x, choose u = eˣ or u = ln x.
如果被积函数含有 eˣ 或 1/x 以及 eˣ 或 ln x 的函数,选择 u = eˣ 或 u = ln x。
Always check that after substitution the integral contains only u and du, with no remaining x terms. If x terms remain, either the substitution is incorrect or further algebraic manipulation is needed.
换元后始终要检查积分中是否只包含 u 和 du,没有剩余的 x 项。如果仍有 x 项,说明换元选择不正确,或者还需要进一步的代数处理。
4. Worked Example: Algebraic Substitution | 例题:代数换元
Integrate ∫ x√(x² + 1) dx. Let u = x² + 1, so du/dx = 2x, hence dx = du/(2x). The integral becomes ∫ x √u × (du/(2x)) = ½ ∫ u^(1/2) du = ½ × (2/3) u^(3/2) + C = ⅓ (x² + 1)^(3/2) + C.
计算 ∫ x√(x² + 1) dx。令 u = x² + 1,则 du/dx = 2x,因此 dx = du/(2x)。积分变为 ∫ x √u × (du/(2x)) = ½ ∫ u^(1/2) du = ½ × (2/3) u^(3/2) + C = ⅓ (x² + 1)^(3/2) + C。
Let u = x² + 1 → du/dx = 2x → dx = du/(2x)
令 u = x² + 1 → du/dx = 2x → dx = du/(2x)
∫ x√(x²+1) dx = ½ ∫ √u du = ⅓ u^(3/2) + C = ⅓ (x²+1)^(3/2) + C
∫ x√(x²+1) dx = ½ ∫ √u du = ⅓ u^(3/2) + C = ⅓ (x²+1)^(3/2) + C
Notice that after replacing dx with du/(2x), the x in the numerator cancels with the x in the denominator. This cancellation is the key sign that the substitution is working correctly.
注意,在把 dx 替换为 du/(2x) 后,分子中的 x 与分母中的 x 相互约掉。这种约分是换元选择正确的重要标志。
5. Definite Integrals and Changing Limits | 定积分与更换积分限
For definite integrals, you have two options: substitute back to x and use the original limits, or change the limits to match the new variable u. If x = a gives u = g(a) and x = b gives u = g(b), then ∫a^b f(g(x)) g'(x) dx = ∫g(a)^g(b) f(u) du. Changing limits often saves time and reduces errors.
对于定积分,你有两种选择:代回 x 并使用原来的积分限,或者把积分限换成新变量 u 的值。如果 x = a 对应 u = g(a),x = b 对应 u = g(b),那么 ∫a^b f(g(x)) g'(x) dx = ∫g(a)^g(b) f(u) du。更换积分限通常能节省时间并减少错误。
Example: evaluate ∫₀¹ 2x (x² + 1)³ dx. Let u = x² + 1. Then du/dx = 2x, so dx = du/(2x). When x = 0, u = 1; when x = 1, u = 2. The integral becomes ∫₁² u³ du = [u⁴/4]₁² = 16/4 – 1/4 = 15/4.
例题:计算 ∫₀¹ 2x (x² + 1)³ dx。令 u = x² + 1。则 du/dx = 2x,所以 dx = du/(2x)。当 x = 0 时,u = 1;当 x = 1 时,u = 2。积分变为 ∫₁² u³ du = [u⁴/4]₁² = 16/4 – 1/4 = 15/4。
∫₀¹ 2x (x² + 1)³ dx = ∫₁² u³ du = [u⁴/4]₁² = 15/4
∫₀¹ 2x (x² + 1)³ dx = ∫₁² u³ du = [u⁴/4]₁² = 15/4
If you choose to substitute back to x instead, you must not use the u-limits with an x-expression. This is one of the most common errors in A-Level exams.
如果你选择代回 x,就不能在 x 表达式中使用 u 的积分限。这是 A-Level 考试中最常见的错误之一。
6. Trigonometric Substitutions | 三角换元
Some integrals require substituting a trigonometric function. For example, ∫ sin² x cos x dx can be solved with u = sin x, du/dx = cos x, so du = cos x dx. The integral becomes ∫ u² du = u³/3 + C = ⅓ sin³ x + C.
有些积分需要把三角函数作为换元变量。例如,∫ sin² x cos x dx 可以令 u = sin x,du/dx = cos x,因此 du = cos x dx。积分变为 ∫ u² du = u³/3 + C = ⅓ sin³ x + C。
∫ sin² x cos x dx = ⅓ sin³ x + C
∫ sin² x cos x dx = ⅓ sin³ x + C
Other useful trigonometric substitutions include u = cos x when sin x appears as an odd power, and u = tan x when sec² x appears. In Edexcel A-Level Maths, you will most commonly use u = sin x, u = cos x, or u = tan x. The key is to match the derivative: du/dx = cos x for u = sin x, du/dx = -sin x for u = cos x, and du/dx = sec² x for u = tan x.
其他有用的三角换元包括:当 sin x 以奇数次幂出现时令 u = cos x;当出现 sec² x 时令 u = tan x。在 Edexcel A-Level 数学中,最常用的是 u = sin x、u = cos x 或 u = tan x。关键是要匹配导数:u = sin x 时 du/dx = cos x;u = cos x 时 du/dx = -sin x;u = tan x 时 du/dx = sec² x。
7. Common Mistakes and How to Avoid Them | 常见错误与避免方法
Many students forget to rewrite dx in terms of du. Always compute du/dx and isolate dx carefully. Another frequent error is mixing x and u in the same integral; the integrand must be entirely in u after substitution. Also, for definite integrals, remember to change the limits or revert to x before applying original limits.
许多学生忘记把 dx 用 du 表示。一定要先计算 du/dx,并小心地分离出 dx。另一个常见错误是在同一个积分中混用 x 和 u;换元后被积函数必须完全用 u 表示。此外,对于定积分,要记得更换积分限,或者在代回 x 后再使用原来的积分限。
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Forgetting to replace dx: always write dx = du / (du/dx) explicitly.
忘记替换 dx:一定要明确写出 dx = du / (du/dx)。
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Leaving x terms after substitution: if x remains, revise your choice of u.
换元后仍留有 x 项:如果 x 仍然存在,说明换元选择需要调整。
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Using u-limits with an x-answer or x-limits with a u-answer: keep the variable consistent.
在 x 表达式中使用 u 的积分限,或在 u 表达式中使用 x 的积分限:要保持变量一致。
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Dropping the constant of integration in indefinite integrals: always add + C.
不定积分漏掉积分常数:始终要加上 + C。
8. Exam-Style Practice Questions | 考试风格练习题
Try these questions to test your understanding of integration by substitution. For each one, identify the substitution u, find du/dx, rewrite the integral, integrate, and substitute back or change limits as appropriate.
尝试以下题目来检验你对换元积分法的理解。每题都要写出换元变量 u,求出 du/dx,重写积分,积分,并根据情况代回原变量或更换积分限。
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Question (i): ∫ x e^(x²) dx. Let u = x², so du/dx = 2x and dx = du/(2x). Answer: ½ e^(x²) + C.
题 (i):∫ x e^(x²) dx。令 u = x²,则 du/dx = 2x,dx = du/(2x)。答案:½ e^(x²) + C。
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Question (ii): ∫ cos³ x sin x dx. Let u = cos x, so du/dx = -sin x and dx = du/(-sin x). Answer: -¼ cos⁴ x + C.
题 (ii):∫ cos³ x sin x dx。令 u = cos x,则 du/dx = -sin x,dx = du/(-sin x)。答案:-¼ cos⁴ x + C。
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Question (iii): ∫₁² x/(x² + 1) dx. Let u = x² + 1, so du/dx = 2x and dx = du/(2x). Limits become u = 2 and u = 5. Answer: ½ ln(5/2).
题 (iii):∫₁² x/(x² + 1) dx。令 u = x² + 1,则 du/dx = 2x,dx = du/(2x)。积分限变为 u = 2 和 u = 5。答案:½ ln(5/2)。
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Question (iv): ∫ 2x (x² – 3)⁵ dx. Let u =
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