📚 Exercise 24C: Integration by Substitution Techniques | 换元积分技巧练习24C
Integration by substitution, often the first advanced technique you encounter after mastering basic antiderivatives, is central to Exercise 24C. This set of problems trains you to recognise when a simple change of variable can transform a messy integral into a standard form. Developing fluency in substitution is essential for IB Mathematics, whether you are taking Analysis and Approaches or Applications and Interpretation at Standard or Higher Level.
换元积分法，通常是掌握基本原函数后遇到的第一个高级技巧，是练习24C的核心。这组题目训练你在什么情况下一个简单的变量代换就能把复杂的积分转化成标准形式。无论你在IB数学中学习的是分析与方法还是应用与解释，标准课程还是高阶课程，熟练掌握换元法都是不可或缺的。
1. The Core Idea of Substitution | 换元法的核心思想
Substitution reverses the chain rule for differentiation. If an integrand contains a composite function multiplied by (or closely related to) the derivative of the inner function, you can set the inner function as a new variable u.
换元法是微分的链式法则的逆过程。如果被积函数包含一个复合函数，并且乘以（或紧密关联于）内层函数的导数，就可以把内层函数设为一个新变量 u。
The goal is always to rewrite the integral entirely in terms of u, so that the du part naturally absorbs the remaining differential and the integration becomes straightforward.
目标始终是把积分完全用 u 表达，使得 du 部分自然地吸收剩下的微分，积分变得简单直接。
2. Recognising a Substitution Opportunity | 识别换元的机会
he derivative of x² is 2x. Setting u = x² gives du = 2x dx, matching the factor perfectly.
寻找一个函数，其导数也作为因子出现在被积函数中。例如，在 ∫ 2x·cos(x²) dx 中，x² 的导数是 2x。设 u = x² 就得到 du = 2x dx，完美匹配。
, are also common when the integrand involves an awkward linear term raised to a power or inside a trigonometric, exponential, or logarithmic function.
当被积函数含有带幂次的线性项，或者三角函数、指数函数、对数函数内部是一个笨拙的线性项时，也经常做线性换元，即令 u = ax + b。
3. The Standard Procedure in Five Steps | 五步标准流程
e u. Identify an inner function g(x) whose derivative appears nearby. Write down u = g(x).
步骤一：选取 u。找到一个内层函数 g(x)，其导数附近出现。写下 u = g(x)。
Step 2: Differentiate to find du/dx, then express dx in terms of du: dx = du / g'(x). This step should ultimately link dx and du elegantly.
步骤二：求导得到 du/dx，然后把 dx 用 du 表示：dx = du / g'(x)。这一步最后要让 dx 和 du 优雅地建立联系。
Step 3: Substitute everything. Replace all x-expressions with u equivalents, including any leftover x's, which must be converted using your original substitution.
步骤三：代入所有内容。把所有 x 的表达式用 u 的等价形式替换掉，包括任何剩余的 x 必须用最初设的 u 表达式转换。
Step 4: Integrate with respect to u. Now you should have a standard integral in u alone. Compute it.
步骤四：对 u 积分。此时你应该得到一个纯粹关于 u 的标准积分。计算它。
Step 5: Back-substitute. Replace u with g(x) to return to the original variable x. Add the constant of integration +C.
步骤五：代回原变量。用 g(x) 换回 u，恢复变量 x。加上积分常数 +C。
4. Worked Example 1: Polynomial Power | 例题一：多项式的幂
Consider ∫ 3x² (x³ + 1)⁵ dx. Let u = x³ + 1, so du = 3x² dx. The integral becomes ∫ u⁵ du.
考虑 ∫ 3x² (x³ + 1)⁵ dx。设 u = x³ + 1，于是 du = 3x² dx。积分变成 ∫ u⁵ du。
∫ u⁵ du = ⅙ u⁶ + C
Back-substituting gives the final answer: ⅙ (x³ + 1)⁶ + C. Notice how the factor 3x² was absorbed completely by du.
代回变量得到最终答案：⅙ (x³ + 1)⁶ + C。注意系数 3x² 是如何完全被 du 吸收的。
5. Worked Example 2: Exponential Function Substitution | 例题二：指数函数换元
Integrate ∫ e2x+1 = 2x+1, then du = 2 dx, so dx = ½ du. The integral transforms to ∫ eᵘ · ½ du = ½ ∫ eᵘ du.
积分 ∫ e2x+1 dx。设 u = 2x+1，则 du = 2 dx，所以 dx = ½ du。积分变换为 ∫ eᵘ · ½ du = ½ ∫ eᵘ du。
½ eᵘ + C → ½ e2x+1 + C
This example highlights linear substitution, where dx becomes a constant multiple of du rather than matching derivatives exactly.
这个例子强调了线性换元，dx 变成了 du 的常数倍，而不是导数完全匹配的情形。
6. Worked Example 3: Trigonometric Substitution | 例题三：三角函数换元
x cos x dx. Set u = sin x, so du = cos x dx. The integrand sin³x · cos x dx becomes u³ du.
求解 ∫ sin³x cos x dx。设 u = sin x，则 du = cos x dx。被积函数 sin³x · cos x dx 变为 u³ du。
∫ u³ du = ¼ u⁴ + C = ¼ sin⁴x + C
Even when odd powers are present, one factor of cos x (or sin x) often serves as the derivative needed for substitution.
即使出现了奇次幂，一个 cos x（或 sin x）的因子常常正好充当换元所需的导数。
7. Substitution with Radicals: An Important Trick | 含根式的换元：一个重要技巧
For ∫ x√x+1 dx, let u = x+1. Then x = u – 1 and du = dx. The integral becomes ∫ (u–1)√u du = ∫ (u3/2 – u1/2) du.
对于 ∫ x√x+1 dx，令 u = x+1。那么 x = u – 1，且 du = dx。积分变成 ∫ (u–1)√u du = ∫ (u3/2 – u1/2) du。
∫ u3/2 du – ∫ u1/2 du = (2/5)u5/2 – (2/3)u3/2 + C
Back-substituting, we obtain (2/5)(x+1)5/2 – (2/3)(x+1)3/2 + C. This demonstrates that substitution can also simplify radicals that are not inside a pure derivative match.
代回变量，得到 (2/5)(x+1)5/2 – (2/3)(x+1)3/2 + C。这表明换元法也可以简化那些并非完美导数匹配的根式。
8. Handling Definite Integrals with Substitution | 用换元法处理定积分
When dealing with a definite integral, you have two options: either find the indefinite integral first and then evaluate using the original limits, or change the limits along with the substitution.
处理定积分时，你有两种选择：要么先求出不定积分再用原始上下限计算，要么在做换元的同时改变积分限。
1. Then du = 2x dx. When x = 0, u = 1; when x = 1, u = 2. The integral becomes ∫₁² 1/u du.
计算 ∫₀¹ 2x / (x²+1) dx。令 u = x²+1，则 du = 2x dx。当 x = 0 时 u = 1；当 x = 1 时 u = 2。积分变成 ∫₁² 1/u du。
[ln|u|]₁² = ln 2 – ln 1 = ln 2
Changing the limits can save time and reduce back-substitution errors, especially in an exam context.
改变积分限可以节省时间，并减少代回变量时出错的概率，尤其是在考试环境中。
9. Common Mistakes in Exercise 24C and How to Avoid Them | 练习24C中的常见错误及避免方法
stake 1: Forgetting to replace dx completely. After choosing u, always compute dx = du / u' and substitute all x-terms.
错误一：忘记完全替换 dx。选定 u 后，一定要计算 dx = du / u'，并替换掉所有的 x 项。
(x) and there is still an extra x factor that is not absorbed, you must express that x in terms of u from the original substitution equation.
错误二：留下一个多余的 x。如果 u = g(x) 且仍有没被吸收的 x 因子，必须根据最初设定的 u = g(x) 把那个 x 用 u 表示。
Mistake 3: Incorrect constant multiple. With linear substitutions, the dx replacement often introduces a factor like 1/a; forgetting this leads to a wrong answer.
错误三：常系数错误。在线性换元中，dx 的替换通常会引入一个如 1/a 的因子；忘记这一点会导致答案错误。
Carefully checking each step against the chain rule derivative is a reliable way to catch such slip-ups.
对照链式法则的求导过程仔细检查每一步，是发现这类疏漏的可靠方法。
10. IB Exam Strategy for Substitution Questions | 换元问题的IB考试策略
In Paper 1 (no calculator), substitution questions are deliberately structured so that the derivative of your chosen u will almost always appear explicitly or be easily manipulated. If the problem feels impossibly messy, reconsider your choice of u.
在试卷一（无计算器）中，换元题目是精心设计的，你选择的 u 的导数几乎总是明显可见或易于处理。如果题目看起来乱成一团，请重新考虑你对 u 的选择。
On Paper 2, you may encounter substitution where the inner function’s derivative is not an exact match but can be created by multiplying or dividing by a constant. Be prepared to introduce adjusting constants and account for them in the final answer.
在试卷二中，你可能会遇到内层函数的导数不完全匹配，但可以通过乘以或除以常数制造出来的换元。要准备好引入调整常数并在最终答案中体现出来。
Showing clear substitution steps, including the explicit rewriting of dx and any limit changes, earns valuable method marks even if the final simplification contains a minor error.
清楚地展示换元步骤，包括明确写出 dx 的转换以及积分限的改变，即使最终化简有小错，也能获得宝贵的过程分。
11. Key Takeaways from Exercise 24C | 练习24C的核心要点
Substitution is a mechanical process once you identify the inner function. Practice with polynomial, exponential, logarithmic, and trigonometric integrands sharpens your pattern recognition.
一旦你能识别内层函数，换元就是一个机械过程。通过练习多项式、指数、对数和三角函数的被积形式，可以磨练你的模式识别能力。
Always verify your result by differentiating it. If the derivative returns the original integrand (up to a constant factor), your substitution was successful.
始终通过求导来验证结果。如果导数能得到原始的被积函数（相差常数倍），那么你的换元就是成功的。
Remember that some integrals require a combination of substitution with other techniques such as integration by parts or partial fractions, but for Exercise 24C a well-chosen u is usually sufficient.
请记住，有些积分需要将换元法与其他技巧如分部积分法或部分分式结合使用，但在练习24C中，精心选择u通常足以解决问题。
12. Final Practice Challenge | 最终练习挑战
x) dx, so you need to compensate with a factor of ⅓. Check your answer by differentiating.
尝试积分 ∫ cos(3x) · e^(sin(3x)) dx，令 u = sin(3x)。注意到 du = 3 cos(3x) dx，所以需要用因子 ⅓ 补偿。通过求导检查你的答案。
If you can handle this without hesitation, you are fully ready for any substitution problem on your IB exam. Keep revisiting Exercise 24C until every variant feels familiar.
如果你能毫不犹豫地解出此题，你就完全为IB考试中任何换元问题做好了准备。反复重温练习24C，直到每一种变化都感到熟悉。
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