Exercise 14C: Integration by Substitution | 练习14C：代换积分法

📚 Exercise 14C: Integration by Substitution | 练习14C：代换积分法

Integration by substitution is one of the most powerful techniques in IB Mathematics: Analysis and Approaches, and it is often tested in both Paper 1 and Paper 2. It reverses the chain rule for differentiation and allows us to evaluate complicated integrals by introducing a new variable. Exercise 14C typically provides practice on this method, moving from simple polynomial antiderivatives to integrals involving trigonometric, exponential and rational functions.

代换积分法是 IB 数学分析与方法课程中最有力的技巧之一，常在试卷一和试卷二中出现。它反转了微分的链式法则，通过引入一个新变量来计算复杂的积分。练习14C通常针对这一方法进行训练，从简单的多项式反导数逐步过渡到含有三角函数、指数函数和有理函数的积分。

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1. Understanding the Chain Rule in Reverse | 理解反链式法则

Integration by substitution works backwards: if we spot an integrand that can be expressed as f'(g(x)) · g'(x), we can let u = g(x) and the integral becomes ∫ f'(u) du, which is simply f(u) + C.

当我们对复合函数如 sin(3x + 1) 求导时，会将外层函数的导数乘以内层函数的导数：d/dx [sin(3x+1)] = 3 cos(3x+1)。代换积分法逆向而行：如果被积函数可以写成 f'(g(x)) · g'(x) 的形式，令 u = g(x)，积分就变成 ∫ f'(u) du，结果就是 f(u) + C。

In IB syllabi, you are expected to recognise integrals of the form ∫ f(g(x))·g' gonometric functions, as long as the derivative of the inner function appears as a factor.

在 IB 教学大纲中，你需要会识别形如 ∫ f(g(x))·g'(x) dx 的积分并执行代换 u = g(x)。该方法不仅适用于多项式，也适用于指数函数、对数函数和三角函数，只要内层函数的导数作为一个因子出现即可。

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2. The Substitution Method: Step by Step | 代换法：逐步解析

Let u = g(x). Then differentiate to find du/dx = g'(x), which gives du = g'(x) dx. Replace every occurrence of g(x) with u, and g'(x) dx with du. The original integral ∫ f(g(x))·g' to u and substitute back u = g(x) to express the answer in terms of the original variable.

令 u = g(x)。然后求导得到 du/dx = g'(x)，即 du = g'(x) dx。把所有的 g(x) 替换成 u，把 g'(x) dx 替换成 du。原来的积分 ∫ f(g(x))·g'(x) dx 就变成了 ∫ f(u) du。最后对 u 积分，再回代 u = g(x)，用原始变量表示答案。

Consider the integral ∫ 2x·(x² + 3)⁴ dx. Here g(x) = x² + 3, and g'x) = 2x, which appears exactly. Set u = x² + 3, du = 2x dx. The integral becomes ∫ u⁴ du = ⅕ u⁵ + C. Substituting back gives ⅕ (x² + 3)⁵ + C.

考虑积分 ∫ 2x·(x² + 3)⁴ dx。这里 g(x) = x² + 3，g'(x) = 2x 恰好出现。设 u = x² + 3，du = 2x dx。积分变为 ∫ u⁴ du = ⅕ u⁵ + C。回代后得到 ⅕ (x² + 3)⁵ + C。

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3. Choosing the Right Substitution u | 选择合适的代换变量 u

Successful substitution depends on identifying a suitable u. Usually u is the 'inner function' pressions, pick u = the expression inside the bracket. For exponential functions such as e^ax+b u see sin(kx) or cos(kx), often u = kx works, especially when the derivative k is present.

成功的代换取决于找到合适的 u。通常 u 是“内层函数”，其导数出现在被积函数的其他部分。对于多项式表达式，选取括号内的表达式作为 u。对于指数函数如 e^ax+b，令 u = ax+b。对于三角函数，如果看到 sin(kx) 或 cos(kx)，常常可以令 u = kx，尤其当导数 k 存在时效果更好。

iting tan x = sin x / cos x and letting u = cos x, du = -sin x dx leads to ∫ -1/u du = -ln|u| single rule works for all cases, which is why Exercise 14C provides a variety of families.

有时需要发挥创造力。对于 ∫ tan x dx，把 tan x 写成 sin x / cos x，令 u = cos x，du = -sin x dx 可导出 ∫ -1/u du = -ln|u| + C。没有一条规则能适用于所有情形，这正是练习14C提供多种类型积分的原因。

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4. Example 1: Basic Polynomial Substitution | 示例1：基本多项式代换

du = 6x² dx. The integral becomes ∫ u⁷ du = ⅛ u⁸ + C. Replacing u gives ⅛ (2x³ - 5)⁸ + C.

计算 ∫ 6x² (2x³ - 5)⁷ dx。令 u = 2x³ - 5，则 du/dx = 6x²，即 du = 6x² dx。积分变为 ∫ u⁷ du = ⅛ u⁸ + C。代回 u 得 ⅛ (2x³ - 5)⁸ + C。

Always check your answer by differentiation: the derivative of ⅛ (2x³ - 5)⁸ is 6x² (2x³ - 5)⁷, confirming correctness. This routine is strongly recommended in IB exams to avoid simple algebraic errors.

务必用求导检验答案：对 ⅛ (2x³ - 5)⁸ 求导得 6x² (2x³ - 5)⁷，确认正确。IB 考试中强烈推荐这一习惯，以避免简单的代数错误。

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5. Example 2: Trigonometric Substitution | 示例2：三角函数代换

x = du/5. The integral becomes ∫ sin u · (1/5) du = -1/5 cos u + C = -1/5 cos(5x + 1) + C.

计算 ∫ sin(5x + 1) dx。由于 5x + 1 的导数是 5，这个 5 虽然缺失，我们仍然可用 u = 5x + 1，du = 5 dx，因此 dx = du/5。积分变为 ∫ sin u · (1/5) du = -1/5 cos u + C = -1/5 cos(5x + 1) + C。

2x dx, so x dx = du / 2. The integral becomes ½ ∫ cos u du = ½ sin u + C = ½ sin(x²) + C.

当准确的导数没有出现时，常常需要进行常数因子的调整。在 ∫ x cos(x²) dx 中，设 u = x²，du = 2x dx，因此 x dx = du / 2。积分变为 ½ ∫ cos u du = ½ sin u + C = ½ sin(x²) + C。

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6. Example 3: Exponential and Logarithmic Functions | 示例3：指数与对数函数

For ∫ e^4x dx, let u = 4x, du = 4 dx, so dx = du/4. The integral becomes ¼ ∫ e^u du = ¼ e^u + C = ¼ e^4x + C. If a factor x accompanies e^x², as in ∫ 3x e^x² dx, let u = x², du = 2x dx, so 3x dx = (3/2) du; thus the integral equals (3/2) e^u + C = (3/2) e^x² + C.

对于 ∫ e^4x dx，令 u = 4x，du = 4 dx，故 dx = du/4。积分变为 ¼ ∫ e^u du = ¼ e^u + C = ¼ e^4x + C。如果因子 x 与 e^x² 同时出现，如 ∫ 3x e^x² dx，设 u = x²，du = 2x dx，于是 3x dx = (3/2) du；因此积分等于 (3/2) e^u + C = (3/2) e^x² + C。

Rational integrals of the form ∫ (f'(x)/f(x)) dx directly give ln|f(x)| nstance, ∫ (2x/(x²+9)) dx: set u = x²+9, du = 2x dx, so the integral becomes ∫ 1/u du = ln|u| + C = ln(x²+9) + C. Recognising this pattern will save time in the exam.

形如 ∫ (f'(x)/f(x)) dx 的有理积分直接得到 ln|f(x)| + C。例如 ∫ (2x/(x²+9)) dx：设 u = x²+9，du = 2x dx，积分成为 ∫ 1/u du = ln|u| + C = ln(x²+9) + C。识别这一模式能在考试中节省时间。

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7. Substitution with Definite Integrals | 定积分的代换

When dealing with definite integrals, you have two options: either integrate indefinitely, substitute back, and then apply the original limits, or change the limits at the moment of substitution. The second method is usually safer and faster because you never need to revert to the original variable.

处理定积分时有两种选择：可以先求不定积分，代回原变量再代入原始上下限；也可以在代换的同时变换积分限。第二种方法通常更安全、更快速，因为你不再需要回到原变量。

Given ∫₀¹ x = 0, u = 1; when x = 1, u = 2. Also du = 2x dx, so 4x dx = 2 du. The integral becomes ∫₁² 2 u³ du = 2 [u⁴/4]₁² ½ (16 - 1) = 15/2. Notice how substituting the limits made the evaluation straightforward.

给定 ∫₀¹ 4x (x²+1)³ dx，令 u = x²+1。当 x = 0 时 u = 1；当 x = 1 时 u = 2。同时 du = 2x dx，因此 4x dx = 2 du。积分变为 ∫₁² 2 u³ du = 2 [u⁴/4]₁² = ½ (16 - 1) = 15/2。注意变换上下限如何使计算变得简洁。

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8. Changing Limits of Integration Correctly | 正确变换积分限

d will yield an incorrect result. As soon as you set u = g(x), calculate the new limits: u_lower = g(a), u_upper = g(b). Write them next to the integral sign.

许多 IB 学生在用代换法计算定积分时忘记更新积分限。把原积分限留在 u-积分上是严重失误，会得出错误结果。一旦设 u = g(x)，立即计算新上下限：u_lower = g(a)，u_upper = g(b)，并写在积分号旁。

For ∫_π/6^π/3 = π/6, u = π/3; when x = π/3, u = 2π/3. du = 2 dx, so dx = du/2. The integral becomes ½ ∫_π/3^2π/3 sin u du = -½[cos u]_π/3^2π/3 = -½(-½ - ½) = ½. Check this with the direct antiderivative to build confidence.

对于 ∫_π/6^π/3 sin(2x) dx，设 u = 2x。上下限：x = π/6 时 u = π/3；x = π/3 时 u = 2π/3。du = 2 dx，故 dx = du/2。积分变为 ½ ∫_π/3^2π/3 sin u du = -½[cos u]_π/3^2π/3 = -½(-½ - ½) = ½。你可以用直接求反导数的方法检验，以增强信心。

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9. Common Mistakes to Avoid | 常见错误及避免方法

One frequent mistake is failing to express the entire integrand in terms of u. If an x factor remains, the integration cannot proceed. Ensure that after the substitution, all x's are replaced by expressions in u, or that the remaining x is cancelled by the du replacement.

一个常见错误是未能将被积函数完全用 u 表示。如果留下一个 x 因子，积分无法继续。确保代换后所有 x 都被替换成关于 u 的表达式，或者剩下的 x 被 du 的替换消去。

Another error is forgetting the constant factor adjustment. When du = k·g'(x) dx but only g' dx, so dx = du/7, giving a factor 1/7. Missing this 1/7 will change the result completely.

另一个错误是忘记常量因子调整。当 du = k·g'(x) dx 但被积函数中只出现 g'(x) dx 时，必须除以 k。例如 ∫ cos(7x) dx，若取 u = 7x，du = 7 dx，故 dx = du/7，会得到因子 1/7。漏掉这个 1/7 会完全改变结果。

With definite integrals, students often apply the original limits to the u-integral without converting them. Always ask: "What is u when x equals the lower limit? What is u when x equals the upper limit?" Write these new values clearly.

对于定积分，学生经常不转换积分限就直接将原上下限用于 u-积分。要始终问自己：“当 x 等于下限时 u 是多少？当 x 等于上限时 u 是多少？”清晰写出这些新值。

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10. Practice Tips for IB Exam Success | IB 考试成功练习技巧

In Paper 1, where no calculator is allowed, substitution problems are designed to have neat arithmetic. Practise the standard families: polynomial brackets, exponential compositions, trigonometric arguments with linear inner functions, and rational functions where the numerator is the derivative of the denominator.

在试卷一中不能使用计算器，代换题目都设计得数字工整。练习标准类型：多项式括号、指数复合、内层为线性函数的三角函数，以及分子为分母导数的有理函数。

During revision, always attempt to predict a good u before reading the solution. Use a drill method: cover the answer, set u, find du, rewrite the integral, integrate, then check

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