Integration by Substitution | 换元积分法

📚 Integration by Substitution | 换元积分法

Integration by substitution is a powerful technique that reverses the chain rule for differentiation. In the IB Mathematics Analysis and Approaches course, both SL and HL students must master this method to handle integrals of composite functions. By choosing an appropriate substitution u = g(x), we transform a complicated integral into a simpler one that can be evaluated directly. This article explains the theory, provides worked examples, and highlights common pitfalls to help you succeed in IB exams.

换元积分法是一种强大的技巧，它逆转了微分的链式法则。在 IB 数学分析与方法课程中，无论是 SL 还是 HL 的学生都必须掌握这一方法，才能处理复合函数的积分。通过选择合适的代换 u = g(x)，我们可以把一个复杂的积分转化为可以直接计算的较简单积分。本文阐释理论、提供例题并指出常见错误，帮助你在 IB 考试中取得成功。

1. What is Integration by Substitution? | 什么是换元积分法？

Integration by substitution is essentially the reverse process of the chain rule for differentiation. When we differentiate a composite function like sin(3x² + 1), we multiply by the derivative of the inner function: cos(3x² + 1) × 6x. Substitution allows us to integrate expressions where a function and its derivative appear together, effectively “undoing” the chain rule.

换元积分法本质上是微分链式法则的逆过程。当我们对复合函数如 sin(3x² + 1) 求导时，需要乘以内层函数的导数：cos(3x² + 1) × 6x。换元法让我们能够对被积函数中含有一个函数及其导数的表达式进行积分，从而有效“撤销”链式法则。

2. The Substitution Rule (Indefinite Integrals) | 不定积分的换元法则

The rule states: if u = g(x) is differentiable and f is continuous on the range of g, then ∫ f(g(x))·g'(x) dx = ∫ f(u) du. In practice, we set u to be a suitable inner function, compute du = g'(x) dx, and rewrite the whole integral in terms of u. After integrating, we substitute back to x.

法则为：如果 u = g(x) 可微，且 f 在 g 的值域上连续，那么 ∫ f(g(x))·g'(x) dx = ∫ f(u) du。实际操作中，我们将 u 设为合适的内层函数，计算 du = g'(x) dx，并用以 u 表示整个被积表达式。积分后再将 u 换回 x。

∫ f(g(x))·g'(x) dx = ∫ f(u) du, where u = g(x)

∫ f(g(x))·g'(x) dx = ∫ f(u) du，其中 u = g(x)

For example, to integrate ∫ 2x·cos(x²) dx, we notice that the derivative of x² is 2x. Let u = x², then du = 2x dx. The integral becomes ∫ cos(u) du = sin(u) + C = sin(x²) + C.

例如，积分 ∫ 2x·cos(x²) dx，我们注意到 x² 的导数是 2x。令 u = x²，则 du = 2x dx。积分变为 ∫ cos(u) du = sin(u) + C = sin(x²) + C。

3. How to Choose u | 如何选取 u

Choosing the right u is crucial. Generally, let u be the “inside” function of a composition, or an expression whose derivative also appears (up to a constant factor) in the integral. Common choices include: u = expression inside a bracket like (3x+5)⁴, u = the angle in a trigonometric function, u = the exponent in an exponential function, or u = the denominator of a rational function when its derivative sits in the numerator.

选取正确的 u 至关重要。通常，令 u 为复合函数中的“内层”函数，或者令 u 为导数（可能差一个常数因子）也出现在被积函数中的表达式。常见选择包括：u = 括号内的表达式，如 (3x+5)⁴；u = 三角函数的角度部分；u = 指数函数的指数部分；或 u = 有理函数的分母，且其导数出现在分子中。

Always compute du and check whether the remaining x-dependence can be completely replaced by u and constants. If not, you may need a different substitution or a combination with algebraic manipulation.

始终要计算 du 并检查剩余的关于 x 的部分能否完全被 u 和常数替代。如果不能，可能需要换另一种代换，或结合代数变形。

4. Basic Worked Examples | 基础例题

Example 1: Evaluate ∫ (2x + 3)√(x² + 3x − 5) dx. Let u = x² + 3x − 5. Then du = (2x + 3) dx. The integral becomes ∫ √u du = ∫ u^(1/2) du = (2/3)u^(3/2) + C = (2/3)(x² + 3x − 5)^(3/2) + C.

例 1：计算 ∫ (2x + 3)√(x² + 3x − 5) dx。令 u = x² + 3x − 5，则 du = (2x + 3) dx。积分变为 ∫ √u du = ∫ u^(1/2) du = (2/3)u^(3/2) + C = (2/3)(x² + 3x − 5)^(3/2) + C。

Example 2: Evaluate ∫ e^(sin x) cos x dx. Let u = sin x, du = cos x dx. The integral simplifies to ∫ e^u du = e^u + C = e^(sin x) + C.

例 2：计算 ∫ e^(sin x) cos x dx。令 u = sin x，du = cos x dx。积分简化为 ∫ e^u du = e^u + C = e^(sin x) + C。

Example 3: Evaluate ∫ (ln x)/x dx. Set u = ln x, then du = (1/x) dx. The integral becomes ∫ u du = u²/2 + C = (ln x)²/2 + C.

例 3：计算 ∫ (ln x)/x dx。设 u = ln x，则 du = (1/x) dx。积分变成 ∫ u du = u²/2 + C = (ln x)²/2 + C。

5. Substitution with Definite Integrals | 定积分的换元法

For definite integrals, substitution requires a change in the limits of integration. If we substitute u = g(x), then the limits x = a and x = b become u = g(a) and u = g(b). The integral ∫ₐᵇ f(g(x))·g'(x) dx equals ∫_{g(a)}^{g(b)} f(u) du. This method avoids the need to substitute back to x and simplifies evaluation.

对于定积分，换元法需要同时改变积分限。如果令 u = g(x)，那么积分限 x = a 和 x = b 就变成 u = g(a) 和 u = g(b)。积分 ∫ₐᵇ f(g(x))·g'(x) dx 等于 ∫_{g(a)}^{g(b)} f(u) du。这种方法省去了换回 x 的步骤，简化了计算。

Example: Evaluate ∫₀¹ 2x·(x² + 1)³ dx. Let u = x² + 1, du = 2x dx. When x = 0, u = 1; when x = 1, u = 2. The integral becomes ∫₁² u³ du = [u⁴/4]₁² = (16/4 − 1/4) = 15/4.

例题：计算 ∫₀¹ 2x·(x² + 1)³ dx。令 u = x² + 1，du = 2x dx。当 x = 0 时 u = 1；当 x = 1 时 u = 2。积分变为 ∫₁² u³ du = [u⁴/4]₁² = (16/4 − 1/4) = 15/4。

6. Changing Limits of Integration | 积分限的变换

It is critical to transform the limits correctly when using definite integrals. Always write the new limits in terms of u, using the substitution equation. Mistakes often occur when students keep the original x-limits while integrating with respect to u. You may also choose to find the antiderivative in terms of u, substitute back to x, and then use the original limits, but this is longer. The transformed-limits approach is cleaner and recommended in IB exams.

在计算定积分时，正确变换积分限至关重要。务必用代换方程将新的积分限表示为 u 的值。学生们常犯的错误是在对 u 积分时仍保留原来的 x 积分限。你也可以先求出关于 u 的原函数，再换回 x 并代入原积分限，但这种方法更麻烦。变换积分限的方法更简洁，也是 IB 考试中推荐的做法。

Be careful: if the substitution u = g(x) is not monotonic on the interval, the limits transformation may need extra care, but typical IB problems involve continuous, monotonic substitutions.

注意：如果代换 u = g(x) 在积分区间上不是单调的，变换积分限可能需要特别处理，不过典型的 IB 题目都涉及连续且单调的代换。

7. Trigonometric Substitution | 三角换元

Trigonometric substitution is used for integrals involving square roots of quadratic expressions, such as √(a² − x²), √(a² + x²), or √(x² − a²). The idea is to use Pythagorean identities to eliminate the square root. Common substitutions include: for √(a² − x²), let x = a sin θ, then dx = a cos θ dθ and √(a² − x²) = a cos θ. For √(a² + x²), use x = a tan θ. For √(x² − a²), use x = a sec θ.

三角换元用于含有二次根式的积分，例如 √(a² − x²)、√(a² + x²) 或 √(x² − a²)。其思想是利用勾股恒等式消除平方根。常用的代换包括：对 √(a² − x²)，令 x = a sin θ，则 dx = a cos θ dθ，且 √(a² − x²) = a cos θ。对 √(a² + x²)，使用 x = a tan θ。对 √(x² − a²)，使用 x = a sec θ。

Example: Evaluate ∫ dx / √(1 − x²). Let x = sin θ, then dx = cos θ dθ, √(1 − x²) = √(1 − sin²θ) = cos θ. The integral becomes ∫ (cos θ dθ) / cos θ = ∫ dθ = θ + C = arcsin x + C. This matches the standard result.

例题：计算 ∫ dx / √(1 − x²)。令 x = sin θ，则 dx = cos θ dθ，√(1 − x²) = √(1 − sin²θ) = cos θ。积分变为 ∫ (cos θ dθ) / cos θ = ∫ dθ = θ + C = arcsin x + C。这与标准结果一致。

8. Substitution with Inverse Trigonometric Functions | 反三角函数的换元

Integrals leading to inverse trigonometric functions often appear, such as ∫ dx/(a² + x²). Substitution u = x/a simplifies this to (1/a) ∫ du/(1 + u²) = (1/a) arctan u + C. Similarly, patterns like ∫ dx/√(a² − x²) become arcsin(x/a) after a linear substitution. Recognizing these forms is a key skill for IB.

经常出现能积出反三角函数的积分，例如 ∫ dx/(a² + x²)。代换 u = x/a 可将其化简为 (1/a) ∫ du/(1 + u²) = (1/a) arctan u + C。类似地，∫ dx/√(a² − x²) 经过线性代换即可得到 arcsin(x/a)。识别这些形式是 IB 考试的关键技能。

Example: Evaluate ∫ dx/(4x² + 9). Factor 9: = (1/9) ∫ dx/( (2x/3)² + 1 ). Let u = 2x/3, du = (2/3) dx → dx = (3/2) du. Then the integral becomes (1/9)·(3/2) ∫ du/(u²+1) = (1/6) arctan(u) + C = (1/6) arctan(2x/3) + C.

例题：计算 ∫ dx/(4x² + 9)。提取 9：= (1/9) ∫ dx/( (2x/3)² + 1 )。令 u = 2x/3，du = (2/3) dx → dx = (3/2) du。则积分变为 (1/9)·(3/2) ∫ du/(u²+1) = (1/6) arctan(u) + C = (1/6) arctan(2x/3) + C。

9. Recognizing Integrals of the Form f'(x)/f(x) | 识别 f'(x)/f(x) 型积分

A special and highly-tested case is ∫ f'(x)/f(x) dx = ln|f(x)| + C. Here substitution u = f(x) gives du = f'(x) dx, immediately yielding ∫ du/u = ln|u| + C. Examples include ∫ 2x/(x²+1) dx, ∫ cos x/sin x dx, and ∫ 1/(x ln x) dx (with u=ln x). Always check if the numerator is exactly the derivative of the denominator.

一种常见且高频考察的情况是 ∫ f'(x)/f(x) dx = ln|f(x)| + C。这里代换 u = f(x) 得到 du = f'(x) dx，直接得出 ∫ du/u = ln|u| + C。例子包括 ∫ 2x/(x²+1) dx、∫ cos x/sin x dx，以及 ∫ 1/(x ln x) dx（令 u = ln x）。务必要检查分子是否恰好是分母的导数。

Example: ∫ (sec²x)/(tan x) dx. Notice the derivative of tan x is sec²x. Let u = tan x, du = sec²x dx. The integral becomes ∫ du/u = ln|u| + C = ln|tan x| + C.

例题：∫ (sec²x)/(tan x) dx。注意到 tan x 的导数是 sec²x。令 u = tan x，du = sec²x dx。积分变为 ∫ du/u = ln|u| + C = ln|tan x| + C。

10. Common Pitfalls and Tips | 常见错误与技巧

Pitfall 1: Forgetting to replace dx entirely. When you have du = g'(x) dx, you must solve for dx and substitute, not just change the function. Many students write du instead of dx without the proper conversion factor. Always express dx = du / g'(x) if g'(x) is not directly visible in the original integral.

错误 1：忘记完全替换 dx。当你得到 du = g'(x) dx 时，必须解得 dx 并代入，而不能只改变函数。许多学生直接写 du 而缺少正确的转换因子。如果原积分中没有直接出现 g'(x)，一定要写出 dx = du / g'(x)。

Pitfall 2: Mixing variables. In indefinite integrals, after evaluating ∫ f(u) du, always rewrite the result in terms of x. Leaving an answer with u is considered incomplete. For definite integrals, use new limits and never return to x.

错误 2：变量混淆。计算不定积分时，求出 ∫ f(u) du 后，务必将结果用 x 表示。答案中含 u 被认为是不完整的。对于定积分，使用新的积分限且不用再换回 x。

Pitfall 3: Incorrect limits order. When substituting u = g(x), if g(a) > g(b), the new integral will have the upper limit smaller than the lower limit. This is mathematically fine, but be careful when applying the fundamental theorem.

错误 3：积分限顺序错误。代入 u = g(x) 时，如果 g(a) > g(b)，新的积分上限会小于下限。这在数学上是允许的，但在应用微积分基本定理时要小心。

Tip: For rational functions of sin and cos, the Weierstrass substitution t = tan(x/2) can be useful but is beyond typical IB scope unless guided.

提示：对于含有 sin 和 cos 的有理函数，万能代换 t = tan(x/2) 有时会派上用场，但这超出了典型 IB 的范畴，除非题目有引导。

11. Practice Tips for IB Exams | IB 考试备考建议

In IB Math AA exams, substitution questions often combine with other techniques. You might see an integral that requires a substitution after a slight algebraic manipulation, or a definite integral where the geometry interpretation simplifies the work. Practice identifying the “derivative pair” quickly. Use past paper questions and mark schemes to familiarise yourself with the expected presentation.

在 IB 数学 AA 考试中，换元法题目常常与其他技巧结合。你可能会遇到一个积分，需要稍作代数变形后再换元；或者一个定积分，其几何意义可以简化计算。练习快速识别“导数对”。利用历年真题和评分方案，熟悉所期望的书写格式。

Always show the substitution and the computation of du clearly. For definite integrals, state the new limits explicitly. The IB marking criteria reward a logical, well-structured method.

务必清晰展示代换和 du 的计算过程。对于定积分，要明确写出新的积分限。IB 评分标准青睐逻辑清晰、结构良好的解题步骤。

If a question asks for an exact value, leave your answer in simplified exact form, using fractions and surds. Do not approximate unless explicitly required.

如果题目要求精确值，请将答案保留为简化的精确形式，使用分数和根号。除非明确要求，否则不要取近似值。

12. Summary | 总结

Integration by substitution is an essential tool for IB students. The core idea is to transform the integral into a basic form by a clever change of variable. Mastering when and how to substitute—and handling the bounds correctly—will unlock a large family of integrals. Remember to practise recognising patterns such as f'(x)f(x), linear inner functions, and the use of trigonometric substitution for roots of quadratics. With consistent practice, substitution becomes second nature, giving you confidence in Paper 1 and Paper 2.

换元积分法是 IB 学生必备的重要工具。核心思想是通过巧妙的变量代换，将积分转化为基本形式。掌握何时以及如何代换，并正确处理积分限，将能解决一大类积分问题。记得练习识别诸如 f'(x)f(x) 的模式、线性内层函数，以及用三角换元处理二次根式。通过持续练习，换元法将成为你的第二天性，让你在试卷一和试卷二中信心十足。

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Integration by Substitution | 换元积分法