Exercise 15C: Integration by Substitution | 练习15C：换元积分法

📚 Exercise 15C: Integration by Substitution | 练习15C：换元积分法

Exercise 15C in your IB Mathematics coursebook focuses on integration by substitution, a fundamental technique for evaluating indefinite and definite integrals. This method reverses the chain rule of differentiation, allowing us to simplify complex integrals into standard forms. Mastering this skill is essential for success in both Analysis and Approaches and Applications and Interpretation syllabi, particularly for calculus-based problems.

IB数学教材中的练习15C聚焦于换元积分法，这是一种计算不定积分和定积分的基本技巧。该方法通过逆向应用微分的链式法则，将复杂的积分简化为标准形式。掌握这项技能对于在分析与方法和应用与解释两门课程中取得成功至关重要，尤其是对于微积分类问题。

1. Introduction to Exercise 15C | 练习15C简介

Exercise 15C is a dedicated problem set designed to build fluency in recognising when and how to apply the substitution rule. The problems range from straightforward polynomial substitutions to more intricate trigonometric and exponential integrands. By working through this exercise, you will develop the pattern recognition needed to spot a function and its derivative, which is the hallmark of a successful substitution.

练习15C是一套旨在培养判断何时以及如何应用换元法则的熟练度的专项习题集。题目涵盖从简单的多项式换元到更复杂的三角和指数被积函数。通过完成这套练习，你将培养识别函数及其导数的模式识别能力，这是成功进行换元的关键标志。

In IB examinations, integration by substitution is commonly tested in Paper 1 (non‑calculator) and Paper 2 (calculator‑allowed) sections. The technique is not only a standalone topic but also a prerequisite for solving differential equations, finding areas under curves, and working with kinematics problems. A solid grasp of Exercise 15C will therefore pay dividends across the syllabus.

在IB考试中，换元积分法通常在试卷一（不可使用计算器）和试卷二（可使用计算器）部分进行考查。该方法不仅是一个独立主题，也是求解微分方程、计算曲线下方面积以及处理运动学问题的先决条件。因此，扎实掌握练习15C将使你在整个课程大纲中获益。

2. Understanding Integration by Substitution | 理解换元积分法

Integration by substitution is the reverse process of the chain rule. When differentiating a composite function f(g(x)), the chain rule gives f'(g(x)) · g'(x). Therefore, if we encounter an integral of the form ∫ f'(g(x)) · g'(x) dx, we can reverse this by letting u = g(x), so that du = g'(x) dx, transforming the integral into ∫ f'(u) du, which is typically much easier to evaluate.

换元积分法是链式法则的逆过程。当对复合函数 f(g(x)) 求导时，链式法则给出 f'(g(x)) · g'(x)。因此，如果我们遇到形如 ∫ f'(g(x)) · g'(x) dx 的积分，我们可以通过令 u = g(x) 来逆转这一过程，从而 du = g'(x) dx，将积分转化为 ∫ f'(u) du，这样通常更容易计算。

The key idea is to choose a substitution that simplifies the integrand, converting a complicated expression in x into a simpler one in u. After integrating with respect to u, we must substitute back to express the result in terms of the original variable x (for indefinite integrals) or change the limits accordingly (for definite integrals). This decision is critical and separates a well‑executed substitution from a haphazard one.

关键思想是选择一个能简化被积函数的换元，将复杂的含 x 表达式转化为简单的含 u 表达式。在对 u 积分后，我们必须将结果代回原变量 x（对于不定积分），或者相应地改变积分限（对于定积分）。这一决定至关重要，是高质量换元与随意换元的分水岭。

3. The Substitution Rule: Formula and Notation | 换元法则：公式与符号

The general formula for integration by substitution is: ∫ f(g(x)) · g'(x) dx = ∫ f(u) du, where u = g(x) and du = g'(x) dx. For definite integrals, the formula becomes: ∫ₐᵇ f(g(x)) · g'(x) dx = ∫_{g(a)}^{g(b)} f(u) du. Notice how the limits of integration change from x‑values a and b to u‑values g(a) and g(b).

换元积分的一般公式为：∫ f(g(x)) · g'(x) dx = ∫ f(u) du，其中 u = g(x) 且 du = g'(x) dx。对于定积分，公式变为：∫ₐᵇ f(g(x)) · g'(x) dx = ∫_{g(a)}^{g(b)} f(u) du。注意积分限如何从 x 值 a 和 b 变为 u 值 g(a) 和 g(b)。

The notation du/dx is often used to find the differential. Starting from u = g(x), we differentiate to get du/dx = g'(x), which implies du = g'(x) dx. This step is crucial for replacing dx in the original integral. It is sometimes helpful to solve explicitly for dx in terms of du, especially when the substitution is not a simple polynomial or trigonometric function.

通常使用 du/dx 记法来求微分。从 u = g(x) 出发，求导得 du/dx = g'(x)，这蕴含着 du = g'(x) dx。这一步对于替换原积分中的 dx 至关重要。有时明确解出 dx 用 du 表示会很有帮助，尤其是当换元并非简单的多项式或三角函数时。

4. Step-by-Step Guide for Substitution | 换元积分步骤详解

Step 1: Identify a suitable substitution. Look for an inner function u = g(x) whose derivative g'(x) also appears in the integrand. Step 2: Compute the derivative du/dx and write du = g'(x) dx. Step 3: Rewrite the entire integral in terms of u and du, ensuring absolutely no x terms remain. If some x persists, express it in terms of u using the substitution equation. Step 4: Evaluate the new integral with respect to u. Step 5: For indefinite integrals, substitute back u = g(x) to express the answer in x. For definite integrals, change the limits to u‑values and evaluate directly.

步骤一：确定合适的换元。寻找内函数 u = g(x)，其导数 g'(x) 也出现在被积函数中。步骤二：计算导数 du/dx 并写出 du = g'(x) dx。步骤三：将整个积分用 u 和 du 重写，确保绝对没有剩余的 x 项。如果仍有 x 残留，请使用换元等式将其用 u 表示。步骤四：计算关于 u 的新积分。步骤五：对于不定积分，将 u = g(x) 代回，用 x 表示答案。对于定积分，将积分限改为 u 值并直接计算。

Always check that after substitution the integrand becomes a standard form – a power, exponential, trigonometric, or rational function – that we can integrate using basic rules. Sometimes algebraic manipulation, such as multiplying and dividing by a constant, is needed before the substitution works perfectly. This preliminary adjustment is a frequent source of mistakes, so double‑check your work.

始终检查换元后被积函数是否变成了标准形式——幂函数、指数函数、三角函数或有理函数——从而可以使用基本规则进行积分。有时在换元完美工作之前需要进行代数处理，例如乘以和除以一个常数。这种初步调整是常见的错误来源，因此请仔细检查你的步骤。

5. Example 1: Simple Polynomial Substitution | 例1：简单多项式换元

Evaluate ∫ 2x · (x² + 1)³ dx. Let u = x² + 1. Then du/dx = 2x, so du = 2x dx. The integral immediately becomes ∫ u³ du. This integrates to u⁴/4 + C. Substituting back gives (x² + 1)⁴/4 + C.

计算 ∫ 2x · (x² + 1)³ dx。令 u = x² + 1。则 du/dx = 2x，故 du = 2x dx。积分立即变为 ∫ u³ du。积分得 u⁴/4 + C。代回得 (x² + 1)⁴/4 + C。

This example showcases the ideal situation where the derivative of the inner function appears exactly as a factor. If the constant factor is not an exact match, say ∫ x (x²+1)³ dx, we can rewrite it as (1/2) ∫ 2x (x²+1)³ dx and then proceed with the same substitution u = x²+1. The final answer would be (1/2)(x²+1)⁴/4 + C = (x²+1)⁴/8 + C. This technique of inserting a missing constant is used repeatedly throughout Exercise 15C.

此例展示了理想情况：内函数的导数恰好作为因子出现。如果常数因子不完全匹配，例如 ∫ x (x²+1)³ dx，我们可以将其重写为 (1/2) ∫ 2x (x²+1)³ dx，然后进行同样的换元 u = x²+1。最终答案为 (1/2)(x²+1)⁴/4 + C = (x²+1)⁴/8 + C。这种补足缺失常数的技巧在练习15C中反复使用。

6. Example 2: Integrating Functions with Linear Inside | 例2：含有线性内函数的积分

Evaluate ∫ e^(3x+2) dx. Let u = 3x + 2. Then du/dx = 3, so du = 3 dx, hence dx = du/3. The integral becomes ∫ e^u · (du/3) = (1/3) e^u + C. Substituting back yields (1/3) e^(3x+2) + C.

计算 ∫ e^(3x+2) dx。令 u = 3x + 2。则 du/dx = 3，故 du = 3 dx，因此 dx = du/3。积分变为 ∫ e^u · (du/3) = (1/3) e^u + C。代回得 (1/3) e^(3x+2) + C。

Similarly, for integrals of the type ∫ sin(ax+b) dx, we set u = ax+b, giving du = a dx, so dx = du/a. The integral becomes ∫ sin u · (du/a) = −(1/a) cos u + C. This pattern can be generalised: ∫ f(ax+b) dx = (1/a) F(ax+b) + C, where F is an antiderivative of f. In your IB exercise set, you will encounter many variations, including cos(5x−3) and e^(−2x+1).

类似地，对于形如 ∫ sin(ax+b) dx 的积分，令 u = ax+b，得 du = a dx，故 dx = du/a。积分变为 ∫ sin u · (du/a) = −(1/a) cos u + C。这种模式可归纳为：∫ f(ax+b) dx = (1/a) F(ax+b) + C，其中 F 是 f 的一个原函数。在你的 IB 练习集中，你会遇到许多变体，包括 cos(5x−3) 和 e^(−2x+1)。

7. Example 3: Trigonometric Substitution | 例3：三角换元

Evaluate ∫ sin³ x cos x dx. Let u = sin x. Then du = cos x dx. The integral becomes ∫ u³ du = u⁴/4 + C = (sin⁴ x)/4 + C. This illustrates the classic choice of the function that is raised to a power as u, with its derivative clearly present.

计算 ∫ sin³ x cos x dx。令 u = sin x。则 du = cos x dx。积分变为 ∫ u³ du = u⁴/4 + C = (sin⁴ x)/4 + C。这展示了经典的选择：将升幂的函数作为 u，其导数明确存在。

Another common trigonometric substitution is found in ∫ tan x dx. Rewrite it as ∫ (sin x / cos x) dx. Set u = cos x, then du = − sin x dx, so −du = sin x dx. The integral turns into ∫ −du/u = −ln|u| + C = −ln|cos x| + C, which is equivalent to ln|sec x| + C. This clever substitution is standard for IB exams and appears in Exercise 15C problems involving combinations of sine and cosine.

8. Example 4: Definite Integrals and Changing Limits | 例4：定积分与换限

Compute ∫₀¹ 2x √(x²+1) dx. Set u = x²+1. Then du = 2x dx. When x = 0, u = 0²+1 = 1; when x = 1, u = 1²+1 = 2. The integral becomes ∫₁² √u du. This evaluates to [ (2/3) u^(3/2) ]₁² = (2/3)(2^(3/2) − 1^(3/2)) = (2/3)(2√2 − 1).

计算 ∫₀¹ 2x √(x²+1) dx。令 u = x²+1，则 du = 2x dx。当 x = 0 时，u = 0²+1 = 1；当 x = 1 时，u = 1²+1 = 2。积分变为 ∫₁² √u du。计算得 [ (2/3) u^(3/2) ]₁² = (2/3)(2^(3/2) − 1^(3/2)) = (2/3)(2√2 − 1)。

The crucial step here is updating the limits of integration. You must compute the new u‑values corresponding to the original x‑limits. You do not need to substitute back to x after integration; evaluating the antiderivative directly in u at the new limits gives the correct answer. Alternatively, you could find the indefinite integral in u, revert to x, and then

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Exercise 15C: Integration by Substitution | 练习15C：换元积分法