Integration by Substitution | 换元积分法

📚 Integration by Substitution | 换元积分法

Integration by substitution is a core technique in A-Level Mathematics that transforms complicated integrals into simpler, standard forms by changing the variable of integration. It is often introduced as the reverse of the chain rule for differentiation and is essential for tackling a wide range of functions encountered in the Edexcel Pure Mathematics syllabus, including polynomial, trigonometric, exponential, and rational integrands.

换元积分法是 A-Level 数学中的一项核心技术，它通过改变积分变量，将复杂的积分转化为更简单、标准的形式。该方法通常被作为微分链式法则的逆运算引入，对于处理 Edexcel 纯数学大纲中遇到的各种函数——包括多项式、三角、指数和有理函数——至关重要。

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1. Introduction to Integration by Substitution | 换元积分法简介

The basic idea of substitution is to replace a complicated expression inside an integral with a single new variable, commonly denoted as u. This simplifies the integrand and often makes the integration step immediate. Once the integral is evaluated in terms of u, the result can be rewritten in the original variable, or, in the case of definite integrals, the limits can be changed to avoid returning to the original variable.

代换的基本思想是用一个新的变量（通常记为 u）替换积分内的复杂表达式。这会简化被积函数，通常使积分步骤变得直接。在用 u 计算出积分后，可以把结果重写回原变量；对于定积分，还可以直接变换积分限，从而避免返回到原变量。

Formally, if we set u = g(x), then the differential du is given by du = g'(x) dx. The original integral ∫ f(g(x)) g' ne question and within larger problems involving area, volumes, or differential equations.

从形式上，如果我们设 u = g(x)，那么微分 du 由 du = g'(x) dx 给出。原来的积分 ∫ f(g(x)) g'(x) dx 就转化为 ∫ f(u) du，通常是一个标准积分。这种方法在 Edexcel 试卷中被大量考查，既可能作为独立问题出现，也会出现在涉及面积、体积或微分方程的大型题目中。

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2. The Reverse Chain Rule | 反向链式法则

The chain rule for differentiation states that d/dx [F(g(x))] = F'(g(x)) · g'(x). Reversing this process gives the fundamental formula for substitution:

微分的链式法则指出 d/dx [F(g(x))] = F'(g(x)) · g'(x)。反向运用这个过程就得到了换元积分的基本公式：

∫ F'(g(x)) g'(x) dx = F(g(x)) + C

the inner function (up to a constant factor). Let u = g(x), then the integral becomes ∫ F'u) du = F(u) + C. The constant factor can be adjusted by manipulating the differential du.

在使用代换时，我们识别出被积函数包含一个复合函数乘以内层函数的导数（可能相差一个常数倍）。令 u = g(x)，则该积分变为 ∫ F'(u) du = F(u) + C。常数因子可以通过调整微分 du 来处理。

For example, consider ∫ 3x² sin(x³) dx. The derivative of x³ is 3x², so we can set u = x³, giving du = 3x² dx. The integral simplifies to ∫ sin u du = -cos u + C = -cos(x³) + C. Always verify your answer by differentiating – it should return the original integrand.

例如，考虑 ∫ 3x² sin(x³) dx。x³ 的导数是 3x²，所以可以令 u = x³，得到 du = 3x² dx。积分简化为 ∫ sin u du = -cos u + C = -cos(x³) + C。务必通过求导来验证答案——它应该能还原出原被积函数。

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3. Identifying the Inside Function u | 识别内层函数 u

Choosing the correct substitution is the key to success. Look for an 'inner function' whose derivative also appears in the integrand (possibly scaled by a constant). Common candidates include:

选择正确的代换是成功的关键。要寻找一个“内层函数”，其导数也出现在被积函数中（可能仅相差一个常数倍）。常见的选择包括：

· Expressions inside brackets raised to a power, e.g., (2x+3)⁵ → u = 2x+3
· The exponent of e, e.g., e^x² → u = x²
· The argument of a trigonometric function, e.g., sin(3x²) → u = 3x²
· The denominator of a rational function, e.g., 1/(x²+4) → u = x²+4
· The logarithm argument, e.g., ln(2x) → u = 2x

· 括号内带有幂次的表达式，如 (2x+3)⁵ → u = 2x+3
· e 的指数部分，如 e^x² → u = x²
· 三角函数的变量，如 sin(3x²) → u = 3x²
· 有理函数的分母，如 1/(x²+4) → u = x²+4
· 对数的变量，如 ln(2x) → u = 2x

Once u is chosen, compute du/dx, solve for dx, and rewrite the entire integral in terms of u. Don't forget to replace every x – if any x remains after substitution, you have likely chosen the wrong u or need to express x in terms of u using the substitution equation.

选定 u 以后，计算 du/dx，解出 dx，并把整个积分用 u 重写。别忘了替换掉所有的 x——如果代换后还剩下任何 x，那么你可能选错了 u，或者需要利用代换方程把 x 用 u 表示出来。

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4. Substitution for Definite Integrals | 定积分的代换

When the integral has limits, you have two options. The first is to treat it as an indefinite integral, substitute back to x after integration, and then apply the original limits. The second, more efficient method is to change the limits of integration to correspond to the new variable u.

当积分带有上下限时，你有两种选择。第一种是当作不定积分处理，积分后再代回 x，然后代入原来的积分限。第二种更高效的方法是把积分限转换为新变量 u 对应的值。

x = b, then the new limits are u = g(a) and u = g(b). The definite integral transforms as:

如果 u = g(x)，原积分限为 x = a 和 x = b，那么新的积分限就是 u = g(a) 和 u = g(b)。定积分变换为：

∫_a^b f(g(x)) g'(x) dx = ∫_g(a)^g(b) f(u) du

This approach saves time and reduces the risk of algebraic mistakes when resubstituting. In Edexcel exams, you must clearly show the change of limits when using this method, stating the new u-values explicitly.

这种方法可以节省时间，并降低在回代时出现代数错误的风险。在 Edexcel 考试中，若使用该方法，你必须清楚地展示积分限的转换，明确写出新的 u 值。

For example, to evaluate ∫₀¹ 2x e^x² dx, let u = x². Then du = 2x dx. When x = 0, u = 0; when x = 1, u = 1. The integral becomes ∫₀¹ e^u du = [e^u]₀¹ = e - 1.

例如，计算 ∫₀¹ 2x e^x² dx，令 u = x²，则 du = 2x dx。当 x = 0 时，u = 0；当 x = 1 时，u = 1。积分变为 ∫₀¹ e^u du = [e^u]₀¹ = e - 1。

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5. Trigonometric Substitutions | 三角代换

nvolving powers of trigonometric functions and their products. A standard pattern is ∫ sinⁿ where the derivative of sin x is cos x, so u = sin x works perfectly. Similarly, for ∫ tanⁿx sec²x dx, let u = tan x because d(tan x)/dx = sec²x.

代换法对于涉及三角函数幂次及其乘积的积分极其有用。一个标准模式是 ∫ sinⁿx cos x dx，其中 sin x 的导数是 cos x，所以令 u = sin x 效果极佳。类似地，对于 ∫ tanⁿx sec²x dx，可以令 u = tan x，因为 d(tan x)/dx = sec²x。

More advanced substitutions involving identities can also be used, such as replacing sin² h the differential: for example, if u = sin 2x, then du = 2 cos 2x dx, not just cos 2x dx.

也可以使用涉及恒等式的更高级代换，例如用 (1 - cos 2x)/2 替换 sin²x。不过，Edexcel 大纲主要考查三角函数的直接 u 代换。请始终留意微分：例如，若 u = sin 2x，则 du = 2 cos 2x dx，而不只是 cos 2x dx。

A typical exam question might ask you to integrate ∫ sin³ tting u = sin θ gives du = cos θ dθ; limits become u = 0 and u = 1. The integral becomes ∫₀¹ u³ du = [u⁴/4]₀¹ = 1/4.

一个典型的考题可能要求你计算 ∫₀^π/2 sin³θ cos θ dθ。令 u = sin θ 得到 du = cos θ dθ；积分限变为 u = 0 和 u = 1。积分变为 ∫₀¹ u³ du = [u⁴/4]₀¹ = 1/4。

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6. Integrating Rational Functions with Substitution | 有理函数的换元积分

tor is (a multiple of) the derivative of the denominator. For an integral of the form ∫ (f'(x)/f(x)) dx, the natural substitution u = f(x) leads directly to ∫ (1/u) du = ln |u| + C.

当分子是分母的导数（或其常数倍）时，有理函数往往可以通过代换来求解。对于形如 ∫ (f'(x)/f(x)) dx 的积分，很自然的代换 u = f(x) 直接给出 ∫ (1/u) du = ln |u| + C。

Consider ∫ (2x + 3)/(x² 3x + 4) dx. The derivative of the denominator is 2x + 3, exactly the numerator. Let u = x² + 3x + 4, so du = (2x + 3) dx. The integral becomes ∫ du/u = ln |u| + C = ln |x² + 3x + 4| + C.

考虑 ∫ (2x + 3)/(x² + 3x + 4) dx。分母的导数是 2x + 3，恰好是分子。令 u = x² + 3x + 4，则 du = (2x + 3) dx。积分变为 ∫ du/u = ln |u| + C = ln |x² + 3x + 4| + C。

the derivative, a linear adjustment constant can often be introduced. For example, ∫ x/(x²+1) dx. Let u = x²+1, du = 2x dx, so x dx = ¹/₂ du. The integral becomes ½ ∫ du/u = ½ ln |u| + C.

如果分子并非恰好为导数，通常可以引入一个线性调整常数。例如，∫ x/(x²+1) dx。令 u = x²+1，du = 2x dx，因此 x dx = ¹/₂ du。积分变为 ½ ∫ du/u = ½ ln |u| + C。

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7. Substitution with Exponentials and Logarithms | 指数与对数形式的代换

Exponential and logarithmic integrals also respond well to substitution. For ∫ e^ax+b re interesting are integrals where the function and its derivative are paired, such as ∫ e^x cos(e^x) dx, where u = e^x gives du = e^x dx.

指数和对数积分也能很好地用代换处理。对于 ∫ e^ax+b dx，简单的线性代换 u = ax + b 可以直接解决。更有趣的是函数与其导数成对出现的积分，例如 ∫ e^x cos(e^x) dx，此时令 u = e^x 得到 du = e^x dx。

larly, integrals involving ln x can be simplified by setting u = ln x, which implies x = e^u and dx = e^u clearly benefits from u = ln x because du = (1/x) dx, reducing the integral to ∫ u du = u²/2 + C.

同样，涉及 ln x 的积分可以通过令 u = ln x 来简化，这意味着 x = e^u 且 dx = e^u du。例如，∫ (ln x)/x dx 显然适合令 u = ln x，因为 du = (1/x) dx，将积分简化为 ∫ u du = u²/2 + C。

In Edexcel exams, you may see

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