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Integration by Substitution | 换元积分法详解

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📚 Integration by Substitution | 换元积分法详解

Integration by substitution is a powerful and widely used technique for evaluating integrals. It is particularly helpful when the integrand contains a composite function and its derivative, and it is often referred to as “u-substitution” because we introduce a new variable u.

换元积分法是求积分的一种重要且常用的方法。当被积函数中包含复合函数及其导数时，这种方法尤其有效。由于我们常引入一个新变量 u，因此又称“u 换元法”。

1. The Reverse Chain Rule | 反链式法则

The chain rule for differentiation states that if F is a function of u and u = g(x), then d/dx [F(g(x))] = F′(g(x)) g′(x). This is one of the most important rules in calculus because it connects the derivative of a composite function with the derivatives of its parts.

微分中的链式法则指出：若 F 是 u 的函数，且 u = g(x)，则 d/dx [F(g(x))] = F′(g(x)) g′(x)。这是微积分中最重要的一条法则，因为它将复合函数的导数与其内部函数的导数联系起来。

If we know that F′(u) = f(u), then the chain rule tells us that F(g(x)) is an antiderivative of f(g(x)) g′(x). Therefore, integration by substitution is the reverse process of the chain rule, which is why it is sometimes called the “reverse chain rule”.

如果已知 F′(u) = f(u)，那么链式法则说明 F(g(x)) 是 f(g(x)) g′(x) 的一个原函数。因此，换元积分法是链式法则的逆过程，所以有时也称为“反链式法则”。

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

In the formula above, the derivative g′(x) explicitly appears in the integrand. When this happens, we can replace g(x) by u and g′(x) dx by du to obtain a simpler integral in terms of u.

在上面的公式中，被积函数中明确出现了导数 g′(x)。当出现这种情况时，我们可以将 g(x) 替换为 u，将 g′(x) dx 替换为 du，从而得到一个关于 u 的更简单的积分。

2. Basic Substitution Procedure | 基本换元步骤

The standard substitution method can be organised into five clear steps. Following these steps will help you solve many integrals systematically.

标准的换元法可以分为五个清晰的步骤。按照这些步骤操作，可以帮助你系统性地解决许多积分问题。

  • Step 1: Choose a substitution u = g(x) that makes the integrand simpler.

    步骤 1：选择一个换元 u = g(x)，使得被积函数更简单。

  • Step 2: Differentiate to find du = g′(x) dx.

    步骤 2：求微分得到 du = g′(x) dx。

  • Step 3: Rewrite the entire integral in terms of u and du, removing all x expressions.

    步骤 3：将整个积分改写为关于 u 和 du 的形式，去掉所有含 x 的表达式。

  • Step 4: Integrate with respect to u.

    步骤 4：对 u 进行积分。

  • Step 5: Substitute back u = g(x) to express the final answer in terms of x, unless the limits have already been changed for a definite integral.

    步骤 5：将 u = g(x) 代回，用 x 表示最终答案；若为定积分且已换限，则无需代回。

This procedure works for both indefinite and definite integrals, but definite integrals require extra care with the limits of integration.

这一过程对不定积分和定积分均适用，但定积分需要额外注意积分上下限的处理。

3. How to Choose the Substitution | 如何选择换元

Choosing the right substitution is the most important skill in this method. There is no universal rule, but a reliable guide is to look for a function whose derivative already appears in the integrand, up to a constant factor.

选择合适的换元是该方法中最关键的技巧。虽然不存在通用法则，但一个可靠的指导原则是：在被积函数中寻找一个函数，其导数已经出现（最多相差一个常数因子）。

For example, in the integral ∫ 2x e^(x²) dx, we notice that the derivative of x² is 2x, which appears as a factor. Therefore, we set u = x² and du = 2x dx, reducing the integral to ∫ e^u du.

例如，在积分 ∫ 2x e^(x²) dx 中，我们发现 x² 的导数为 2x，且 2x 作为因子出现在被积函数中。因此令 u = x²，du = 2x dx，积分就化为 ∫ e^u du。

Another common strategy is to choose u as the denominator when the numerator is its derivative, or to choose u as the exponent of an exponential function. In general, you should experiment and simplify the integrand before deciding on a substitution.

另一个常见策略是：当分子是分母的导数时，令 u 为分母；或者令 u 为指数函数中的指数部分。一般来说，你应该先进行尝试和化简，再决定如何换元。

4. Definite Integrals: Changing the Limits | 定积分：换限

When evaluating a definite integral using substitution, you must also change the limits of integration. Suppose the original integral is ∫ₐᵇ f(g(x)) g′(x) dx. Let u = g(x). Then the new limits are u = g(a) and u = g(b).

用换元法计算定积分时，必须同时改变积分上下限。假设原积分为 ∫ₐᵇ f(g(x)) g′(x) dx，令 u = g(x)，则新限为 u = g(a) 和 u = g(b)。

∫ₐᵇ f(g(x)) g′(x) dx = ∫_{g(a)}^{g(b)} f(u) du

In this way, you never need to substitute back to x after integrating. However, if you prefer, you can leave the antiderivative in terms of x and use the original limits. Both methods are correct, but changing the limits is usually more direct.

这样，积分后就不需要再代回 x。当然，如果你愿意，也可以先求出关于 x 的原函数，再代入原上下限。两种方法都是正确的，但换限通常更直接。

For example, consider ∫₀¹ 2x e^(x²) dx. Let u = x². Then du = 2x dx, and the limits change from x = 0 to u = 0, from x = 1 to u = 1. The integral becomes ∫₀¹ e^u du = e^u evaluated from 0 to 1, which equals e − 1.

例如，考虑 ∫₀¹ 2x e^(x²) dx。令 u = x²，则 du = 2x dx，上限从 x = 1 变为 u = 1，下限从 x = 0 变为 u = 0。积分变为 ∫₀¹ e^u du，即 e^u 在 0 到 1 上的值，结果为 e − 1。

5. Trigonometric Substitutions | 三角换元

Trigonometric substitution is a special type of substitution used when the integrand contains expressions such as √(a² − x²), √(a² + x²), or √(x² − a²). These forms suggest that a trigonometric identity may simplify the square root.

三角换元是一种特殊的换元法，用于被积函数中含有诸如 √(a² − x²)、√(a² + x²) 或 √(x² − a²) 这类表达式的情形。这些形式提示我们，利用三角恒等式可能可以化简根号。

  • For √(a² − x²), use x = a sin θ, so that √(a² − x²) = a cos θ.

    对于 √(a² − x²)，令 x = a sin θ，则 √(a² − x²) = a cos θ。

  • For √(a² + x²), use x = a tan θ, so that √(a² + x²) = a sec θ.

    对于 √(a² + x²)，令 x = a tan θ，则 √(a² + x²) = a sec θ。

  • For √(x² − a²), use x = a sec θ, so that √(x² − a²) = a tan θ.

    对于 √(x² − a²)，令 x = a sec θ，则 √(x² − a²) = a tan θ。

These choices rely on the Pythagorean identities: sin²θ + cos²θ = 1, tan²θ + 1 = sec²θ, and sec²θ − 1 = tan²θ.

这些选择依赖于勾股恒等式：sin²θ + cos²θ = 1，tan²θ + 1 = sec²θ，以及 sec²θ − 1 = tan²θ。

After integrating with respect to θ, you must convert the result back to x using a right-angled triangle or by using the inverse trigonometric functions.

在关于 θ 完成积分后，你需要通过构造直角三角形或使用反三角函数，将结果代回为 x 的形式。

6. Substitutions for Simple Rational Functions | 简单有理函数的换元

Many simple rational functions can be integrated using a linear substitution. For instance, the integral ∫ 1/(ax + b) dx can be evaluated by letting u = ax + b.

许多简单有理函数可以通过线性换元来积分。例如，积分 ∫ 1/(ax + b) dx 可通过令 u = ax + b 来求解。

If u = ax + b, then du = a dx, so dx = du/a. The integral becomes (1/a) ∫ 1/u du = (1/a) ln|u| + C = (1/a) ln|ax + b| + C.

若 u = ax + b，则 du = a dx，从而 dx = du/a。积分化为 (1/a) ∫ 1/u du = (1/a) ln|u| + C = (1/a) ln|ax + b| + C。

Similarly, for integrals of the form ∫ x/(x² + 1) dx, we recognise that the numerator is related to the derivative of the denominator. Let u = x² + 1, then du = 2x dx, giving (1/2) ln|x² + 1| + C.

类似地，对于形如 ∫ x/(x² + 1) dx 的积分，我们注意到分子与分母的导数有关。令 u = x² + 1，则 du = 2x dx，得到 (1/2) ln|x² + 1| + C。

7. Common Pitfalls and How to Avoid Them | 常见误区与避免方法

Integration by substitution is a straightforward method, but students often make small errors. Recognising these common pitfalls will help you avoid losing marks in exams.

换元积分法本身并不复杂，但学生在解题时经常出现一些小错误。识别这些常见误区，有助于你在考试中避免失分。

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