📚 Exercise 16B.1: Integration by Substitution | 替换积分法练习 16B.1
Integration by substitution is one of the most powerful techniques in the IB Mathematics Analysis and Approaches syllabus. Often considered the reverse of the chain rule for differentiation, this method transforms a complicated integral into a simpler one by introducing a new variable. Exercise 16B.1 is designed to build fluency in recognising when and how to apply substitution, from basic algebraic changes to trigonometric and exponential cases. Mastering these exercises will not only prepare you for Paper 1 and Paper 2 but also deepen your understanding of how integrals behave under transformation.
替换积分法是 IB 数学分析与方法课程中最强大的工具之一。该方法常被视为链式求导法则的逆运算，通过引入新变量将复杂积分化为简单形式。练习 16B.1 旨在训练学生识别何时以及如何进行替换，从基础代数替换到三角函数与指数函数的情形。熟练掌握这些练习不仅能为试卷一和试卷二做好准备，也能加深对积分在变量变换下行为的理解。
1. The Essence of Substitution | 替换法的本质
tive of its inner function. This is exactly the pattern left by the chain rule: if we differentiate f(g(x)), we get f'(g(x)) · g' a derivative of an inner part, we can let u = g(x) and rewrite the integral in terms of u.
替换积分法的核心思想在于识别被积函数可以表示为一个复合函数与其内层函数导数的乘积。这正是链式求导法则留下的模式：对 f(g(x)) 求导得到 f'(g(x)) · g'(x)。因此，当我们看到一个函数乘以某个看起来像内层部分导数的因子时，就可以令 u = g(x)，并以 u 为变量重新写出积分。
The notation du = g' ue. Once the substitution is made, the integral should become a basic form such as ∫ un du, ∫ eu du, or ∫ (1/u) du. The success of the method depends entirely on choosing a u that simplifies the integrand dramatically.
记号 du = g'(x) dx 是这一技巧的关键。一旦完成替换，积分应当化为基本形式，例如 ∫ un du、∫ eu du 或 ∫ (1/u) du。此法的成败完全取决于能否选择一个可以极大简化被积函数的 u。
2. The Substitution Formula in Symbols | 替换公式的符号表达
The formal statement of the substitution rule is:
替换法则的正式表述为：
∫ f(g(x)) · g′(x) dx = ∫ f(u) du, where u = g(x)
antiderivative in u, we must back-substitute u = g(x) to return to the original variable.
如果积分以 ∫ h(g(x)) · g′(x) dx 的形式呈现，我们立即令 u = g(x)，于是 du = g′(x) dx。所得积分为 ∫ h(u) du，通常很容易计算。在求出关于 u 的原函数后，必须回代 u = g(x) 以回归原变量。
For definite integrals the formula also requires changing the limits: if x goes from a to b, then u goes from g(a) to g(b). This avoids the need to return to x before evaluating the final value.
对于定积分，该公式还要求更换上下限：若 x 从 a 变到 b，则 u 从 g(a) 变到 g(b)。这样在求最终值之前不必先回到 x。
3. Step-by-Step Procedure | 逐步操作流程
A systematic approach is crucial when tackling Exercise 16B.1. Follow these steps:
在应对练习 16B.1 时，系统性的方法至关重要。请遵循以下步骤：
- Identify the candidate u: Look for an inner function whose derivative also appears (up to a constant factor) in the integrand.
- 选定候选 u：寻找一个内层函数，其导数（可能相差一个常数因子）也出现在被积函数中。
- Compute du: du = u′ dx. Manipulate the expression so that all x and dx are replaced by u and du.
- 计算 du：对 u 关于 x 求导并写出 du = u′ dx。处理该表达式，使得所有 x 和 dx 都被替换为 u 和 du。
- Rewrite the integral entirely in u: Ensure no x remains. If a stray x remains and cannot be replaced, the substitution may need adjustment.
- 将积分完全以 u 重写：确保没有残留的 x。若有未能替换的 x，则可能需要调整替换方式。
- Integrate with respect to u: Evaluate the simpler integral.
- 对 u 积分：计算该简化后的积分。
- Back-substitute (indefinite) or update limits (definite): Return to the original variable or change limits.
- 回代（不定积分）或更新上下限（定积分）：回归原变量或更换积分限。
In Exercise 16B.1 you will see integrals where the derivative of the chosen u appears almost exactly, sometimes missing only a constant multiple like 2 or -1. That constant can be compensated by multiplying the integral appropriately.
在练习 16B.1 中，你会遇到所选 u 的导数几乎完全匹配，有时仅差一个常数因子如 2 或 -1。这些常数可以通过适当调整积分来补偿。
4. Choosing the Right Substitution | 选择正确的替换
The most common difficulty students face is not the execution of substitution but the initial choice of u. A good heuristic is to pick the "inside" of a composite function, especially one raised to a power, under a root, or inside a trigonometric or exponential function.
学生最常见的困难并非执行替换本身，而是最初选择 u。一个实用的启发式方法是挑选复合函数的“内部”，尤其是作为幂、根号内部或处于三角函数、指数函数内部的部分。
n let u = 2x, as the derivative of 2x is 2, which is a constant multiplier easily handled.
例如，在 ∫ 3x² √(x³ + 1) dx 中，x³ + 1 位于根号内部，其导数 3x² 以因子形式出现。因此令 u = x³ + 1。在 ∫ sin(2x) dx 中，可以令 u = 2x，因为 2x 的导数是 2，容易作为常数乘数处理。
If the integrand contains both a function and its derivative clearly, pick that function as u. If multiple choices are possible, test mentally which one simplifies the integral the most. Experience from Exercise 16B.1 will sharpen this intuition.
若被积函数同时包含一个函数及其导数，应选择该函数作为 u。若多种选择皆有可能，可在脑海中试验哪一种能最大程度简化积分。通过练习 16B.1 积累的经验会加深这种直觉。
5. Common Algebraic Patterns | 常见代数模式
Exercise 16B.1 frequently features algebraic structures that are classic targets for substitution. One pattern is the integral of a linear expression raised to a power, where the derivative is merely a constant: ∫ (ax + b)n dx. Here, let u = ax + b, so du = a dx, and the integral becomes (1/a) ∫ un du.
练习 16B.1 经常出现一些经典的可借助替换处理的代数结构。其一是线性表达式的幂次积分，其导数仅为常数：∫ (ax + b)n dx。此时令 u = ax + b，则 du = a dx，积分变为 (1/a) ∫ un du。
dx suggests u = x² + 1, making du = 2x dx, so the integral becomes ∫ 1/u du = ln|u| + C. Recognising this gives a natural logarithm result.
另一常见情形是分子为分母的导数（或其倍数）的分式。例如 ∫ (2x)/(x² + 1) dx 暗示令 u = x² + 1，得到 du = 2x dx，于是积分化为 ∫ 1/u du = ln|u| + C。识别出这一点即可得出自然对数结果。
es du = 2x dx, and the integral becomes ½ ∫ u&sup4; du, which is straightforward.
还应留意函数与其导数幂次相乘的情形，比如 ∫ x(x² + 3)&sup4; dx。设 u = x² + 3 得到 du = 2x dx，积分化为 ½ ∫ u&sup4; du，十分直接。
6. Substitution with Exponential Functions | 指数函数替换
When the integrand contains an exponential function with a composite exponent, substitution works beautifully. For ∫ ekx dx, simply set u = kx; then du = k dx and ∫ ekx dx = (1/k)ekx + C.
当被积函数含有复合指数的指数函数时，替换法效果极佳。对于 ∫ ekx dx，只需令 u = kx，则 du = k dx，∫ ekx dx = (1/k)ekx + C。
More interesting cases appear in Exercise 16B.1, like ∫ 2x ex² sup2; is 2x, so setting u = x²; gives du = 2x dx. The integral transforms into ∫ eu du = eu + C = ex² + C.
练习 16B.1 中会出现更有趣的情形，如 ∫ 2x ex² dx。注意到 x² 的导数是 2x，因此令 u = x²; 得到 du = 2x dx。积分变换为 ∫ eu du = eu + C = ex² + C。
What if the factor in front does not exactly match the derivative? Consider ∫ x ex² vative of x² is 2x, so we write x dx = ½ du. The integral is then ½ ∫ eu du = ½ ex² + C. Recognising and adjusting for constant multipliers is a key skill in this exercise set.
若前面的因子与导数不完全匹配该怎么办？考虑 ∫ x ex² dx。此时 x² 的导数为 2x，因此我们将 x dx 写为 ½ du。积分即为 ½ ∫ eu du = ½ ex² + C。识别并调整常数乘数是本练习集的关键技能。
7. Trigonometric Substitutions | 三角替换
b, so du = a dx, and the integral yields (1/a) sin(ax+b) or -(1/a) cos(ax+b) accordingly.
三角函数的积分常可通过替换处理。练习 16B.1 中可能遇到 ∫ sin(ax + b) dx 或 ∫ cos(ax + b) dx。令 u = ax + b，则 du = a dx，积分相应得到 (1/a) sin(ax+b) 或 -(1/a) cos(ax+b)。
Another classic pattern involves powers of sin and cos when one of them appears with its derivative. For ∫ sin3x cos x dx, setting u = sin x gives du = cos x dx, transforming the integral into ∫ u3 du. Similarly, ∫ cos5x sin x dx becomes -∫ u5 du with u = cos x, due to the minus sign from the derivative of cos x.
另一经典模式涉及 sin 和 cos 的幂次且其中一方与其导数同时出现。对于 ∫ sin3x cos x dx，设 u = sin x 得到 du = cos x dx，积分化为 ∫ u3 du。类似地，∫ cos5x sin x dx 通过设 u = cos x 化为 -∫ u5 du，因为 cos x 的导数带有负号。
ubstitution like u = 2x or u = cos x. Exercise 16B.1 builds this versatility step by step.
要求更高的问题可能需要先用三角恒等式处理被积函数再进行替换。例如，处理 ∫ sin²x dx 或 ∫ tan x dx 通常依赖于重写表达式，然后应用如 u = 2x 或 u = cos x 的替换。练习 16B.1 将逐步构建这种灵活应对的能力。
8. Substitution for Rational Functions | 有理函数的替换
r substitution. The integral ∫ (f′(x))/(f(x)) dx is immediately recognised as ln|f(x)| + C, which can be derived by setting u = f(x).
有理函数，尤其是分子为分母导数（或其倍数）的有理函数，是替换法的理想对象。积分 ∫ (f′(x))/(f(x)) dx 可立即认作 ln|f(x)| + C，这可通过设 u = f(x) 推出。
u = (3x² + 2) dx, exactly the numerator. The integral simplifies to ∫ 1/u du = ln|u| + C = ln|x³ + 2x + 1| + C.
例如，∫ (3x² + 2)/(x³ + 2x + 1) dx：令 u = x³ + 2x + 1，则 du = (3x² + 2) dx，恰好是分子。积分简化为 ∫ 1/u du = ln|u| + C = ln|x³ + 2x + 1| + C。
More intricate rational functions require completing the square or partial fractions before substitution. Exercise 16B.1 provides a carefully graded sequence: from direct ln-forms to those needing a linear substitution first, such as ∫ (1)/((2x+1)²) dx or ∫ (x)/(x²+4x+3) dx after simplifying the denominator.
更复杂的有理函数可能需要先配方或部分分式，再进行替换。练习 16B.1 提供了精心分层的练习顺序：从直接的 ln 形式到需要先进行线性替换的题目，比如 ∫ (1)/((2x+1)²) dx 或简化分母后的 ∫ (x)/(x²+4x+3) dx。
9. Definite Integrals and Substitution | 定积分与替换
Handling definite integrals with substitution involves one important decision: change limits or back-substitute. Changing the limits as soon as the substitution is made is often more efficient and reduces algebraic errors. The process is: given ∫ab )) g′(x) dx, set u = g(x), transform the limits to g(a) and g(b), and evaluate ∫g(a)g(b) f(u) du.
处理定积分的替换涉及一个重要选择：更换上下限还是先回代。在完成替换的同时即更换上下限通常更高效，并可减少代数错误。做法是：给定 ∫ab f(g(x)) g′(x) dx，令 u = g(x)，将积分限转换为 g(a) 和 g(b)，然后计算 ∫g(a)g(b) f(u) du。
For instance, evaluate ∫01 2x ex² x²; when x = 0, u = 0; when x = 1, u = 1. Also du = 2x dx. The integral becomes ∫01 eu du = [eu]01 = e - 1. Note that we never revert to x, which streamlines the working.
例如，计算 ∫01 2x ex² dx。令 u = x²;；当 x = 0 时 u = 0；当 x = 1 时 u = 1。且 du = 2x dx。积分变为 ∫01 eu du = [eu]01 = e - 1。注意我们完全未回到 x，这使解答更为简洁。
If you choose to back-substitute, you must remember to return the limits to x after integration. Many students forget to apply the original limits to u, leading to incorrect numerical answers. Exercise 16B.1 deliberately mixes problems that best suit the limit-change method, encouraging good habits.
如果选择回代，必须记得在积分后恢复关于 x 的上下限。许多学生忘记将原积分限应用于 u，导致数值答案错误。练习 16B.1 特意混合了最适合更换积分限方法的题目，以鼓励良好习惯。
10. Common Pitfalls and How to Avoid Them | 常见陷阱与规避方法
s dimensionally inconsistent. Always write du = u′ dx and solve for dx if necessary.
即便理论清楚，小失误也可能使完美的替换设置功亏一篑。一个常见错误是忘记彻底替换 dx。如果新积分中仍同时出现 dx 和 du，表达式在量纲上就不一致。务必写出 du = u′ dx，并在必要时解出 dx。
Missing the minus sign will flip the final answer. Double-check the derivative sign before proceeding.
另一陷阱是三角替换中的符号错误，尤其是当 u = cos x 时，因为 du = -sin x dx。遗漏负号会翻转最终答案。在继续之前务必仔细核对导数符号。
Students also occasionally attempt to substitute when the integrand cannot be fully expressed in terms of u. If after substitution there is still an x variable left that cannot be rewritten using u, the chosen u is probably incorrect. In such a case, reconsider the structure or explore a different substitution.
学生偶尔也会在被积函数无法完全用 u 表示时强行替换。若替换后仍残留无法用 u 重写的 x 变量，所选 u 很可能不正确。这时应重新考虑结构或尝试别的替换。
In definite integrals, mixing variables and limits is dangerous. A clear check: once the integral is written in u, the limits must be in u-values. If you see x-values on a u-integral, stop and correct the limits. Exercise 16B.1 includes multiple definite integrals to train this consistency.
在定积分中，混淆变量与积分限非常危险。有一个明确的检验方法：一旦积分用 u 写出，积分限必须是对应 u 的值。如果在一个 u 积分上看到 x 值，应立即停下来纠正上下限。练习 16B.1 包含了多道定积分题，以训练这种一致性。
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