📚 pdfjoiner_(4)-076: Integration by Substitution | 代换积分法
Integration by substitution is one of the most powerful techniques in A‑Level Mathematics. It reverses the chain rule and enables us to integrate composite functions systematically. Mastering this method unlocks a wide range of problems, from simple algebraic integrals to tricky trigonometric ones, and it is a favourite topic in Edexcel exam papers.
代换积分法是A‑Level数学中最强有力的技巧之一。它逆转了链式法则，使我们能够系统地求出复合函数的积分。掌握这一方法可以解锁从简单代数积分到复杂三角积分的大量问题，是爱德思考试中的热门考点。
1. Introduction to Integration by Substitution | 代换积分法简介
When you face an integral that contains a function and its derivative (or a constant multiple of it), substitution is often the key. The idea is to replace a complicated part of the integrand with a new variable u, making the integral much easier to handle. Essentially, you are "un‑doing" the chain rule that would appear in differentiation.
当你遇到一个积分，其中包含一个函数及其导数（或导数的常数倍）时，代换常常是求解的关键。这一思路是用一个新的变量 u 代替被积函数中的复杂部分，从而使积分变得容易处理。从本质上说，你是在“撤销”微分中出现的链式法则。
2. The Chain Rule in Reverse | 反向链式法则
). Then du/dx = g′(x), so du = g′(x) dx. The integral becomes ∫ F′(u) du = F(u) + C. In a more compact form:
回忆一下，若对 y = F(g(x)) 求导，我们得到 y′ = F′(g(x))·g′(x)。因此，∫ F′(g(x))·g′(x) dx = F(g(x)) + C。在代换中，我们设 u = g(x)，那么 du/dx = g′(x)，即 du = g′(x) dx。积分就变为 ∫ F′(u) du = F(u) + C。更紧凑的形式为：
∫ f(g(x))·g′(x) dx = ∫ f(u) du, where u = g(x)
This is the formula you will apply repeatedly. The art lies in recognising the «inner function» g(x) and ensuring that its derivative (or a simple multiple) is also present.
这就是你要反复使用的公式。其中的技巧在于识别出“内层函数”g(x)，并确保它的导数（或其简单倍数）也同时出现。
3. Choosing the Right Substitution | 选择合适的代换
Picking u correctly is the most critical step. Common choices include:
正确挑选 u 是最关键的一步。常见的选择包括：
- The inner function of a composite, e.g. in ∫ (3x+5)⁸ dx, let u = 3x+5.
- 复合函数的内层函数，例如在 ∫ (3x+5)⁸ dx 中，令 u = 3x+5。
- The expression inside a square root, e.g. in ∫ x√x²+1 dx, let u = x²+1.
- 根号内的表达式，例如在 ∫ x√x²+1 dx 中，令 u = x²+1。
- The denominator of a fraction, e.g. in ∫ (2x)/(x²+1) dx, let u = x²+1.
- 分式的分母，例如在 ∫ (2x)/(x²+1) dx 中，令 u = x²+1。
- onometric function that has its derivative present, e.g. in ∫ sin³θ cosθ dθ, let u = sinθ.
- 三角函数内含有导数参与的部分，例如在 ∫ sin³θ cosθ dθ 中，令 u = sinθ。
t multipliers. Remember that substitution works best when the derivative of u appears essentially as a factor in the integrand.
有时你需要调整常数，因为 du = g′(x) dx 可以容纳常数倍数。请记住，当 u 的导数基本上以因子的形式出现在被积函数中时，代换最为有效。
4. Basic Substitution Steps | 基本代换步骤
Follow these steps every time to avoid mistakes.
每次都应遵循以下步骤以避免出错。
- Choose u = g(x).
- 选择 u = g(x)。
- Find du/dx and rearrange to get du = g′(x) dx, or dx = du / g′(x).
- 求出 du/dx，并变形得到 du = g′(x) dx，或 dx = du / g′(x)。
- Substitute all x‑dependencies: replace g(x) by u and dx by du/g′(x). The g′(x) should cancel out.
- 替换所有与 x 相关的部分：将 g(x) 换为 u，将 dx 换为 du/g′(x)。此时 g′(x) 应当被约去。
- Integrate the resulting expression with respect to u.
- 对所得的 u 表达式进行积分。
- Replace u by g(x) to return to the original variable.
- 最后将 u 换回 g(x)，回到原变量。
- Add the constant of integration +C.
- 加上积分常数 +C。
This algorithm works for indefinite integrals. For definite integrals, an extra step modifies the limits, which we will explore shortly.
这一算法适用于不定积分。对于定积分，额外的一步需修改变限，我们稍后会探讨。
5. Example: Linear Substitution | 示例：线性代换
Evaluate ∫ (2x+1)⁵ dx.
计算 ∫ (2x+1)⁵ dx。
/12 + C. Finally, replace u: ((2x+1)⁶)/12 + C. Notice how the constant 2 was simply absorbed by du.
设 u = 2x+1，则 du/dx = 2，于是 dx = du/2。进行代换：∫ (2x+1)⁵ dx = ∫ u⁵·(du/2) = (1/2)∫ u⁵ du = (1/2)·(u⁶/6) + C = u⁶/12 + C。最后换回 u：((2x+1)⁶)/12 + C。注意常数 2 被 du 轻松吸纳了。
6. Handling Definite Integrals | 处理定积分
When working with a definite integral, you can either change the limits along with the substitution or integrate and then back‑substitute before applying the original limits. Changing limits is usually quicker:
在处理定积分时，你可以随着代换一起改变积分上下限，也可以先求原函数再用原变量代回后代入原限。改变限通常更快：
dx, let u = g(x). Then when x = a, u = g(a); when x = b, u = g(b). The integral becomes ∫g(a)g(b) f(u) du. There is no need to return to x.
对于 ∫ₐᵇ f(g(x)) g′(x) dx，令 u = g(x)。那么当 x = a 时，u = g(a)；当 x = b 时，u = g(b)。积分变为 ∫g(a)g(b) f(u) du。无需再回到变量 x。
Example: ∫₀¹ x·√x²+1 = x²+1, so du = 2x dx → x dx = du/2. When x = 0, u = 1; x = 1, u = 2. Integral becomes ∫₁² √u·(du/2) = (1/2)∫₁² u1/2 du = (1/2)·[(2/3)u3/2]₁² = (1/3)(23/2 – 1).
示例：∫₀¹ x·√x²+1 dx。令 u = x²+1，则 du = 2x dx → x dx = du/2。当 x = 0 时，u = 1；x = 1 时，u = 2。积分变为 ∫₁² √u·(du/2) = (1/2)∫₁² u1/2 du = (1/2)·[(2/3)u3/2]₁² = (1/3)(23/2 – 1)。
7. Substitution with Trigonometric Functions | 三角函数代换
Trigonometric integrals are extremely common. The key is to spot a function and its derivative. For example:
三角积分极为常见，关键是识别出一个函数及其导数。例如：
∫ sin³θ cosθ dθ: let u = sinθ, du = cosθ dθ → ∫ u³ du = u⁴/4 + C = (sin⁴θ)/4 + C.
∫ sin³θ cosθ dθ：令 u = sinθ，du = cosθ dθ → ∫ u³ du = u⁴/4 + C = (sin⁴θ)/4 + C。
When integrating powers of tanθ, recall that ∫ tanθ dθ = ln|secθ| itself can be derived via substitution: let u = cosθ, then ∫ tanθ dθ = ∫ (sinθ/cosθ) dθ = –∫ du/u = –ln|u| + C = ln|secθ| + C.
在求 tanθ 的幂次积分时，记住 ∫ tanθ dθ = ln|secθ| + C，这本身可以通过代换导出：令 u = cosθ，则 ∫ tanθ dθ = ∫ (sinθ/cosθ) dθ = –∫ du/u = –ln|u| + C = ln|secθ| + C。
e is obvious, a useful trick is to express everything in terms of sine and cosine and then pick u as one of them.
对于 ∫ sec²θ tanθ dθ，可以试令 u = tanθ，du = sec²θ dθ → ∫ u du = u²/2 + C。如果没有明显的导数，一个有用的技巧是用正弦和余弦表示一切，然后选择 u 为其中之一。
8. Inverse Substitution and Completing the Square | 逆代换与配方法
Sometimes the substitution «u = something in x» is not the best approach. Instead, we let x be a function of a new variable θ – this is called inverse substitution. A classic situation is ∫ √a² – x² dx, where we set x = a sinθ. Then dx = a cosθ dθ and √a² – x² = a cosθ. The integral simplifies to ∫ a² cos²θ dθ, which can be handled using a double‑angle identity.
有时“u = 包含 x 的某个表达式”并不是最好的做法。我们可以反过来，令 x 为新变量 θ 的函数——这称为逆代换。一个经典的情形是 ∫ √a² – x² dx，我们设 x = a sinθ。此时 dx = a cosθ dθ，且 √a² – x² = a cosθ。积分简化为 ∫ a² cos²θ dθ，可利用倍角恒等式处理。
Similarly, for integrals involving √x² + a² or √x² – a², the substitutions x = a tanθ or x = a secθ are used. These are less common in stand‑alone A‑ you often combine with substitution is completing the square under a square root, turning √ax² + bx + c into a form that matches one of these patterns.
类似地，对于包含 √x² + a² 或 √x² – a² 的积分，可以使用 x = a tanθ 或 x = a secθ。这些在独立的A‑Level考试中不常见，但在进阶应用中会出现。另一个常与代换结合的技术是将根号下的式子配方，使 √ax² + bx + c 变成符合上述模式的形式。
9. Common Pitfalls and Tips | 常见陷阱与提示
Even experienced students make mistakes with substitution. Watch out for:
即便是有经验的学生也会在代换中犯错。请留意以下几点：
- Forgetting to change the dx correctly: it is not enough to replace g(x) by u – you must also transform dx using du = g′(x) dx.
- 忘记正确变换 dx：仅仅将 g(x) 替换为 u 是不够的——还必须利用 du = g′(x) dx 转换 dx。
- s off by a constant factor, write dx = (1/k)·du and bring the factor outside the integral.
- 没有调整常数倍数：如果 g′(x) 与积分中的因子相差一个常数倍，应写出 dx = (1/k)·du 并将该常数提到积分外。
- Definite integrals: always update the limits or remember to back‑substitute; mixing both methods often leads to errors.
- 定积分：始终更新积分限，或记住代回后使用原限；将两种方法混用常常导致错误。
- Applying substitution where there is no derivative factor: substitution only works smoothly if the derivative of u appears, possibly after rearranging the integrand.
- 在没有导数因子的地方使用代换：代换仅在 u 的导数出现时（可能需对被积函数稍加变形）才能顺利进行。
- Dropping the +C in indefinite integrals.
- 在不定积分中漏掉 +C。
Always check your answer by differentiating it – this is a fast and reliable way to confirm your work.
务必将你的答案微分验算——这是快速可靠地确认解答的方法。
10. Practice Questions and Solutions | 练习题与解答
Q1: Find ∫ x⁴(x⁵ – 3)⁹ dx.
题1：求 ∫ x⁴(x⁵ – 3)⁹ dx。
Solution: Let u = x⁵ – ⁴ dx → x⁴ dx = du/5. Integral becomes (1/5)∫ u⁹ du = (1/5)·(u¹⁰/10) + C = u¹⁰/50 + C = (x⁵ – 3)¹⁰/50 + C.
解：令 u = x⁵ – 3，则 du = 5x⁴ dx → x⁴ dx = du/5。积分变为 (1/5)∫ u⁹ du = (1/5)·(u¹⁰/10) + C = u¹⁰/50 + C = (x⁵ – 3)¹⁰/50 + C。
Q2: Evaluate the definite integral ∫₀^(π⁄2) cos³x sin x dx.
题2：计算定积分 ∫₀^(π⁄2) cos³x sin x dx。
Solution: Let u = cos x, du = –sin x dx → sin x dx = –du. When x = 0, u = 1; when x = π/2, u = 0. Integral becomes ∫₁⁰ u³ (–du) = ∫₁⁰ –u³ du = [–u⁴/4]₁⁰ = (0) – (–1/4) = 1/4.
解：令 u = cos x，du = –sin x dx → sin x dx = –du。当 x = 0 时，u = 1；当 x = π/2 时，u = 0。积分变为 ∫₁⁰ u³ (–du) = ∫₁⁰ –u³ du = [–u⁴/4]₁⁰ = (0) – (–1/4) = 1/4。
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