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

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

In IB Mathematics, Exercise 15F typically focuses on integration by substitution, a fundamental technique for evaluating both indefinite and definite integrals. This method, often called the reverse chain rule, allows us to transform a complicated integral into a simpler one by introducing a new variable. Mastering substitution is essential for success in the Analysis and Approaches and Applications and Interpretation courses, especially for questions involving trigonometric, exponential, and rational functions.

在 IB 数学中，练习15F通常聚焦于换元积分法，这是计算不定积分和定积分的一项基本技巧。这一方法常被称为链式法则的逆运算，它使我们能够通过引入一个新变量，将复杂的积分转化为更简单的形式。无论是分析数学还是应用数学课程，掌握换元法都是在考试中取得好成绩的关键，尤其是在涉及三角函数、指数函数和有理函数的题目中。

1. The Core Idea of Substitution | 换元法的核心思想

Integration by substitution reverses the chain rule for differentiation. If we have an integral of the form ∫ f(g(x)) · g'(x) dx, we can let u = g(x), so that du = g'(x) dx, and the integral becomes ∫ f(u) du, which may be simpler to evaluate.

换元积分法逆转了微分的链式法则。如果积分具有 ∫ f(g(x)) · g'(x) dx 的形式，我们可以令 u = g(x)，从而 du = g'(x) dx，积分就变成了 ∫ f(u) du，计算起来往往更容易。

The challenge lies in recognizing the inner function and its derivative. In Exercise 15F, you will encounter integrals where the substitution is either given or must be spotted by identifying a function and its derivative appearing together.

难点在于识别内层函数及其导数。在练习15F中，你会遇到一些积分，其中换元方式已经给出，或者需要通过识别一个函数与其导数同时出现来主动发现换元。

2. Choosing the Substitution u | 选择代换变量 u

When the substitution is not provided, a useful rule is to let u be the inner part of a composite function. For example, if the integrand contains sin(3x+1), let u = 3x+1. If it contains e^(x²), try u = x². The differential du must then account for any missing constant factors.

当题目没有给定换元时，一个有用的经验法则是将复合函数的内层设为 u。例如，如果被积函数含有 sin(3x+1)，令 u = 3x+1；如果含有 e^(x²)，则可尝试 u = x²。微分 du 必须弥补缺失的常数因子。

Sometimes the integrand already contains an exact derivative multiple, which makes the substitution straightforward. When the derivative is off by a constant, we can adjust by multiplying and dividing by that constant outside the integral.

有时被积函数已经包含了一个精确的导数倍数，这使得换元变得很简单。当导数只差一个常数倍时，我们可以在积分号外乘除该常数进行调整。

3. The Method Step by Step | 换元积分法步骤

For an indefinite integral:
1. Choose u = g(x) and compute du/dx.
2. Express dx in terms of du: dx = du / g'(x).
3. Substitute u and du into the integral, eliminating all x terms.
4. Integrate with respect to u.
5. Substitute back x into the result.

对于不定积分：
1. 选择 u = g(x) 并计算 du/dx。
2. 用 du 表示 dx：dx = du / g'(x)。
3. 将 u 和 du 代入积分，消去所有含 x 的项。
4. 对 u 积分。
5. 将结果中的 u 换回 x。

For a definite integral ∫ₐᵇ f(g(x))g'(x) dx, we can either change the limits along with the substitution, or integrate without limits and then apply the original x‑limits. Changing limits is faster: compute u(a) and u(b) so that ∫ₐᵇ … dx = ∫_{u(a)}^{u(b)} f(u) du.

对于定积分 ∫ₐᵇ f(g(x))g'(x) dx，我们可以在换元时同时改变积分限，也可以先求原函数再代入原来的 x 限。改变积分限更为快捷：计算 u(a) 和 u(b)，从而得到 ∫ₐᵇ … dx = ∫_{u(a)}^{u(b)} f(u) du。

4. Linear Substitutions | 线性换元

A common type in Exercise 15F involves a linear inner function: ∫ f(ax + b) dx. Let u = ax + b, then dx = du / a. The integral becomes (1/a) ∫ f(u) du. This is a direct application of the reverse chain rule and appears frequently with trigonometric, exponential, and reciprocal functions.

练习15F中常见的一种类型是内层为线性函数的积分：∫ f(ax + b) dx。令 u = ax + b，则 dx = du / a，积分变为 (1/a) ∫ f(u) du。这是逆链式法则的直接应用，经常出现在三角函数、指数函数和倒数函数中。

For example, ∫ sin(2x+1) dx: let u = 2x+1, dx = du/2; ∫ sin u · (du/2) = –(1/2) cos u + C = –(1/2) cos(2x+1) + C.

例如，∫ sin(2x+1) dx：令 u = 2x+1，dx = du/2；∫ sin u · (du/2) = –(1/2) cos u + C = –(1/2) cos(2x+1) + C。

5. Integrals of the Form ∫ f'(x)/f(x) dx | 形如 ∫ f'(x)/f(x) dx 的积分

When the numerator is exactly the derivative of the denominator, the substitution u = f(x) gives du = f'(x) dx, so the integral becomes ∫ du / u = ln|u| + C. Recognizing this pattern saves time.

当分子恰好是分母的导数时，代换 u = f(x) 给出 du = f'(x) dx，于是积分变为 ∫ du / u = ln|u| + C。识别出这种模式能节省大量时间。

For instance, ∫ (2x)/(x²+3) dx: let u = x²+3, du = 2x dx, so the integral is ∫ du/u = ln|u| + C = ln(x²+3) + C (since x²+3 > 0).

例如，∫ (2x)/(x²+3) dx：令 u = x²+3，du = 2x dx，因此积分为 ∫ du/u = ln|u| + C = ln(x²+3) + C（因为 x²+3 > 0）。

6. Substitution with Trigonometric Functions | 涉及三角函数的换元

Integrals like ∫ sin³x cos x dx or ∫ tan x dx respond well to substitution. For ∫ sin³x cos x dx, let u = sin x so du = cos x dx, giving ∫ u³ du. For ∫ tan x dx, write as ∫ (sin x)/(cos x) dx and use u = cos x.

类似 ∫ sin³x cos x dx 或 ∫ tan x dx 的积分很适合用换元法。对于 ∫ sin³x cos x dx，令 u = sin x，则 du = cos x dx，得到 ∫ u³ du。对于 ∫ tan x dx，写成 ∫ (sin x)/(cos x) dx 并令 u = cos x。

Even powers of sine and cosine can sometimes be handled by using double‑angle identities before substitution, but Exercise 15F often separates variables by spotting a function–derivative pair.

正弦和余弦的偶次幂有时可以先使用倍角公式再换元，但练习15F通常是通过发现函数与导数的配对来分离变量。

7. Substitution with Exponential Functions | 涉及指数函数的换元

For integrals involving eᵏˣ where k is a constant, a linear substitution u = kx works. But if the exponent itself is a more complex function, e.g., ∫ x e^(x²) dx, let u = x². Then du = 2x dx, and the integral becomes (1/2) ∫ eᵘ du.

对于包含 eᵏˣ（k 为常数）的积分，线性换元 u = kx 即可。但如果指数本身是一个更复杂的函数，例如 ∫ x e^(x²) dx，则令 u = x²，du = 2x dx，积分变为 (1/2) ∫ eᵘ du。

Similarly, for ∫ aᵏˣ dx, use the fact that aᵏˣ = e^(k ln a · x) and proceed similarly, or directly substitute u = kx and remember the integral of aᵘ.

类似地，对于 ∫ aᵏˣ dx，利用 aᵏˣ = e^(k ln a · x) 并类似处理，或者直接代换 u = kx 并记住 aᵘ 的积分公式。

8. Definite Integrals and Changing Limits | 定积分与积分限的转换

When evaluating a definite integral by substitution, it is often safer to change the limits to the new variable. After substituting u = g(x), compute the new lower limit α = g(a) and upper limit β = g(b). The integral becomes ∫_{α}^{β} f(u) du, and no back‑substitution is needed.

使用换元法计算定积分时，通常更安全的方法是将积分限也转换为新变量的形式。代换 u = g(x) 后，计算新的下限 α = g(a) 和上限 β = g(b)。积分变为 ∫_{α}^{β} f(u) du，无需再换回 x。

For example, ∫₀¹ 2x(x²+1)⁴ dx: let u = x²+1; when x=0, u=1; when x=1, u=2. Then dx = du/(2x), and the integral becomes ∫₁² u⁴ du = [u⁵/5]₁² = 32/5 – 1/5 = 31/5.

例如，∫₀¹ 2x(x²+1)⁴ dx：令 u = x²+1；当 x=0 时 u=1；当 x=1 时 u=2。那么 dx = du/(2x)，积分变为 ∫₁² u⁴ du = [u⁵/5]₁² = 32/5 – 1/5 = 31/5。

9. Spotting Hidden Substitutions | 发现隐藏的换元机会

Sometimes the derivative of the substitution variable is not immediately visible. For instance, in ∫ x √(x+1) dx, letting u = x+1 simplifies the square root, but we must also express x in terms of u: x = u – 1. Then dx = du, and the integral becomes ∫ (u–1)√u du = ∫ (u^(3/2) – u^(1/2)) du, which is straightforward.

有时换元变量的导数并不直接可见。例如，在 ∫ x √(x+1) dx 中，令 u = x+1 能简化根号，但我们也必须用 u 表示 x：x = u – 1。此时 dx = du，积分变为 ∫ (u–1)√u du = ∫ (u^(3/2) – u^(1/2)) du，就可以直接积出了。

This technique of replacing the old variable completely is powerful when the substitution eliminates all awkward parts. Watch for this in the harder questions in Exercise 15F.

当换元能够消除所有棘手的部分时，这种完全替换原变量的技巧非常有效。在练习15F中较难的题目里要留意这一点。

10. Integrating Rational Functions by Substitution | 有理函数的换元积分

Rational functions where the degree of the numerator is one less than that of the denominator suggest a logarithmic substitution. For example, ∫ (x+2)/(x²+4x+7) dx: let u = x²+4x+7, then du = (2x+4) dx = 2(x+2) dx, so the integral is (1/2) ∫ du/u.

当分子次数比分母次数少一次时，有理函数往往提示对数换元。例如，∫ (x+2)/(x²+4x+7) dx：令 u = x²+4x+7，则 du = (2x+4) dx = 2(x+2) dx，因此积分为 (1/2) ∫ du/u。

If the numerator is not an exact multiple, complete the square in the denominator and use a combination of substitution and standard arctan or logarithm forms.

如果分子不是精确的倍数，可先对分母配方，再结合换元与标准的反正切或对数积分公式来求解。

11. Trigonometric Substitutions (Preview for Higher Level) | 三角换元（高水平预备）

While not always required in standard Exercise 15F, trigonometric substitution like x = a sin θ for √(a² – x²) is introduced in Higher Level courses. It transforms radicals into trigonometric forms that are easier to integrate. The substitution x = a tan θ helps with √(a² + x²), and x = a sec θ with √(x² – a²).

虽然标准练习15F不一定会涉及，但在高水平课程中会引入三角换元，例如 x = a sin θ 用于 √(a² – x²)。它将根式转化为更容易积分的三角形式。代换 x = a tan θ 则用于 √(a² + x²)，而 x = a sec θ 用于 √(x² – a²)。

These types of substitutions often feature in IB HL Paper 2 questions and require careful handling of the differential and the limits in definite integrals.

这类换元经常出现在 IB 高水平试卷二的题目中，需要仔细处理微分和定积分的积分限。

12. Common Mistakes and How to Avoid Them | 常见错误及避免方法

1. Forgetting to substitute dx properly: many students write du but leave dx untouched, leading to a mixed‑variable expression. Always replace dx with du / g'(x).
2. Ignoring constant multipliers: if du is off by a constant factor, adjust the integral by multiplying or dividing by that factor.
3. Not changing definite integral limits: this leads to an answer in terms of u evaluated with x values, which is meaningless.
4. Losing the absolute value in ln|u|: remember that the antiderivative of 1/u is ln|u|, not just ln u, unless u is known to be positive.

1. 忘记正确地替换 dx：许多学生写下了 du 却没有动 dx，导致混合变量表达式。务必将 dx 替换为 du / g'(x)。
2. 忽略常数倍数：如果 du 相差一个常数因子，通过乘除该因子来调整积分。
3. 未改变定积分的积分限：这导致用 u 表示的结果却代入 x 的数值，毫无意义。
4. 在 ln|u| 中遗漏绝对值符号：记住 1/u 的原函数是 ln|u|，而不只是 ln u，除非已知 u 为正。

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