Integration by Substitution (Exercise 15G) | 换元积分法（练习15G）

📚 Integration by Substitution (Exercise 15G) | 换元积分法（练习15G）

Integration by substitution is one of the most essential techniques in IB Mathematics, turning complicated integrals into manageable forms by reversing the chain rule. In this article, we will explore the method in depth, illustrate it with examples directly relevant to Exercise 15G, and equip you with the skills needed to tackle any similar problem confidently.

换元积分法是IB数学中最核心的技巧之一，它通过逆向使用链式法则将复杂的积分转化为易于处理的形式。本文将深入探讨这一方法，结合与练习15G直接相关的示例进行讲解，帮助你掌握所需的技能，从容应对各类类似问题。

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

Substitution is essentially the reverse of the chain rule for differentiation. If you have a composite function inside an integral, such as f(g(x)), multiplied by the derivative of the inner function g'(x), you can replace the inner function with a single variable u to simplify the integration.

换元法本质上是微分中链式法则的逆运算。如果被积函数中包含一个复合函数 f(g(x))，并且乘以内部函数 g(x) 的导数 g'(x)，就可以将内部函数替换为一个新变量 u，从而简化积分过程。

The formal statement is: ∫ f(g(x))·g'(x) dx = ∫ f(u) du, where u = g(x) and du = g'(x) dx. This transformation often reduces a messy integral to a basic one that can be integrated directly.

正式的表述为：∫ f(g(x))·g'(x) dx = ∫ f(u) du，其中 u = g(x)，du = g'(x) dx。通过这种变换，繁杂的积分通常能转化为可以直接计算的基本积分。

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2. Step-by-Step Procedure | 分步步骤

To apply the method systematically, follow these steps: (1) Identify a suitable substitution u = g(x) whose derivative also appears in the integrand. (2) Differentiate to find du = g'(x) dx, and solve for dx if necessary. (3) Rewrite the entire integral in terms of u, ensuring no x terms remain. (4) Integrate with respect to u. (5) For indefinite integrals, substitute back u = g(x) to express the result in terms of x; for definite integrals, change the limits of integration to u-values and evaluate directly.

系统地应用该方法可遵循以下步骤：(1) 选择一个合适的替换 u = g(x)，其导数也应大致出现在被积函数中。(2) 求微分得到 du = g'(x) dx，必要时解出 dx。(3) 将整个积分用 u 重新表示，确保不残留 x。(4) 对 u 进行积分。(5) 对于不定积分，将 u = g(x) 回代，用 x 表示结果；对于定积分，需将积分上下限转换为 u 的对应值并直接计算。

Many beginners find it helpful to write out the substitution and the differential on the side: ‘Let u = … , then du = … dx’. This small habit reduces algebraic mistakes.

许多初学者发现，在草稿上明确写出 ‘设 u = …，则 du = … dx’ 这个习惯能有效减少代数错误。

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3. Basic Polynomial Substitutions | 基本多项式换元

Consider the integral ∫ 2x(x² + 1)³ dx. Here the derivative of x² + 1 is 2x, which appears exactly as a factor.

考虑积分 ∫ 2x(x² + 1)³ dx。这里 x² + 1 的导数是 2x，恰好作为一个因子出现。

We set u = x² + 1, then du = 2x dx. The integral becomes ∫ u³ du = u⁴/4 + C. Substituting back gives (x² + 1)⁴/4 + C.

设 u = x² + 1，则 du = 2x dx。积分变为 ∫ u³ du = u⁴/4 + C。回代后得到 (x² + 1)⁴/4 + C。

Another example: ∫ 3x²(x³ − 5)⁴ dx. Choose u = x³ − 5, du = 3x² dx. The integral transforms to ∫ u⁴ du = u⁵/5 + C = (x³ − 5)⁵/5 + C.

另一个例子：∫ 3x²(x³ − 5)⁴ dx。选择 u = x³ − 5，du = 3x² dx。积分变换为 ∫ u⁴ du = u⁵/5 + C = (x³ − 5)⁵/5 + C。

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4. Substitution with Trigonometric Functions | 三角函数的换元

Trigonometric integrals often yield to substitution when a function and its derivative are present. For ∫ sin³x cos x dx, notice that the derivative of sin x is cos x.

当被积函数中同时出现一个三角函数及其导数时，通常可以使用换元法。对于 ∫ sin³x cos x dx，注意到 sin x 的导数是 cos x。

Let u = sin x, then du = cos x dx. The integral becomes ∫ u³ du = u⁴/4 + C = (1/4) sin⁴x + C.

设 u = sin x，则 du = cos x dx。积分变为 ∫ u³ du = u⁴/4 + C = (1/4) sin⁴x + C。

Similarly, ∫ tan x sec² x dx can be integrated by letting u = tan x, du = sec² x dx, giving ∫ u du = u²/2 + C = (1/2) tan² x + C.

类似地，∫ tan x sec² x dx 可以通过设 u = tan x，du = sec² x dx 来求解，得到 ∫ u du = u²/2 + C = (1/2) tan² x + C。

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5. Substitution Involving Exponentials and Logarithms | 涉及指数和对数的换元

When the integrand contains an exponential function whose derivative also appears, substitution simplifies the work. For ∫ eˣ/(1 + eˣ) dx, we see that the derivative of 1 + eˣ is eˣ.

当被积函数包含指数函数且其导数也出现时，换元能简化计算。对于 ∫ eˣ/(1 + eˣ) dx，1 + eˣ 的导数正是 eˣ。

Let u = 1 + eˣ, then du = eˣ dx. The integral reduces to ∫ 1/u du = ln|u| + C = ln(1 + eˣ) + C.

设 u = 1 + eˣ，则 du = eˣ dx。积分化简为 ∫ 1/u du = ln|u| + C = ln(1 + eˣ) + C。

Logarithmic integrals can be handled similarly: ∫ (ln x)/x dx. Set u = ln x, du = (1/x) dx, giving ∫ u du = u²/2 + C = (1/2)(ln x)² + C.

对数积分也可类似处理：∫ (ln x)/x dx。设 u = ln x，du = (1/x) dx，得到 ∫ u du = u²/2 + C = (1/2)(ln x)² + C。

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6. Substitution for Rational Functions | 有理函数的换元

Rational functions where the numerator is a multiple of the derivative of the denominator are perfect candidates for substitution.

当有理函数的分子是分母导数的常数倍时，这恰好是换元法的理想应用场景。

Take ∫ (2x)/(x² + 4) dx. Let u = x² + 4, du = 2x dx. The integral becomes ∫ 1/u du = ln|u| + C = ln(x² + 4) + C, since x² + 4 is always positive.

考虑 ∫ (2x)/(x² + 4) dx。设 u = x² + 4，du = 2x dx。积分变为 ∫ 1/u du = ln|u| + C = ln(x² + 4) + C，因为 x² + 4 恒正。

This pattern ∫ f'(x)/f(x) dx = ln|f(x)| + C is extremely common and can be seen as a special case of substitution.

这种 ∫ f'(x)/f(x) dx = ln|f(x)| + C 的模式极为常见，可以看作是换元法的一个特例。

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7. Definite Integrals: Changing Limits | 定积分：改变积分限

For definite integrals, substitution is even more elegant because you can avoid substituting back to x. Instead, change the limits of integration to correspond to the new variable u.

对于定积分，换元法更加简洁，因为你无需将变量换回 x，只需将积分上下限转换为与新变量 u 对应的值即可。

Evaluate ∫₀¹ 2x(x² + 1)³ dx. Set u = x² + 1. 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.

计算 ∫₀¹ 2x(x² + 1)³ dx。设 u = x² + 1。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。

Remember to convert the limits immediately after deciding on the substitution. Many errors arise from forgetting this step and mistakenly using x-limits with a u-integral.

务必在确定替换后立刻转换上下限。许多错误源于忽略这一步，将 x 的上下限错误地用于 u 的积分。

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8. Recognizing When to Use Substitution | 何时使用换元法的识别

Spotting a substitution is a skill that develops with practice. Look for an ‘inner function’ whose derivative is also present, possibly up to a constant factor.

识别换元是一项需要练习的技能。要寻找一个“内部函数”，其导数也出现在被积函数中，可能只相差一个常数因子。

Typical clues include: a polynomial raised to a power multiplied by its derivative, a trigonometric function composed with another, an exponential expression with its own derivative, or a fraction resembling f'(x)/f(x).

典型线索包括：多项式的幂乘以其导数，一个三角函数与另一个三角函数的复合，指数表达式与其自身的导数，或者形如 f'(x)/f(x) 的分式。

If the integral seems complicated but contains a part whose derivative simplifies the rest, try a substitution. Even when it is not a perfect match, a clever u-substitution often works.

如果积分看起来很复杂，但其中某部分的导数可以简化剩余部分，就可以尝试换元。即使并非完全匹配，巧妙的 u 替换往往也能奏效。

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9. Common Mistakes to Avoid | 常见错误

One frequent error is forgetting to replace dx with du properly. If du = 2x dx, you must solve for dx = du/(2x) and ensure all x’s cancel.

一个常见错误是忘记正确地将 dx 替换为 du。若 du = 2x dx，必须解出 dx = du/(2x)，并确保所有 x 相约去。

Another mistake is leaving the integral in terms of both u and x. After substitution, the whole integrand must be expressed solely in u.

另一个错误是积分中同时出现 u 和 x。换元后，整个被积函数必须完全用 u 表示。

For definite integrals, failing to change the limits is a classic pitfall. Always convert the limits to u-values before integrating, or alternatively, substitute back to x after integration.

对于定积分，未改变上下限是一个经典的陷阱。务必在积分前将上下限转换为 u 值，或者在积分后将结果换回 x。

Finally, do not forget the constant of integration ‘+ C’ for indefinite integrals. IB examiners are strict about this.

最后，不定积分不要忘记加上积分常数 ‘+ C’。IB考官对此要求严格。

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10. Exercise 15G: Practice Problems Explained | 练习15G：习题解析

Let us work through typical problems that could appear in Exercise 15G, reinforcing the technique.

我们一起解答可能出现在练习15G中的典型题目，巩固换元技巧。

Problem 1: ∫ x(2x² + 3)⁵ dx. Choose u = 2x² + 3, so du = 4x dx, thus x dx = du/4. The integral becomes (1/4) ∫ u⁵ du = (1/4)·(u⁶/6) + C = (1/24)(2x² + 3)⁶ + C.

题目1：∫ x(2x² + 3)⁵ dx。设 u = 2x² + 3，则 du = 4x dx，因此 x dx = du/4。积分变为 (1/4) ∫ u⁵ du = (1/4)·(u⁶/6) + C = (1/24)(2x² + 3)⁶ + C。

Problem 2: ∫ cos x sin⁴x dx. Let u = sin x, du = cos x dx. The integral is ∫ u⁴ du = u⁵/5 + C = (1/5) sin⁵x + C.

题目2：∫ cos x sin⁴x dx。设 u = sin x，du = cos x dx。积分为 ∫ u⁴ du = u⁵/5 + C = (1/5) sin⁵x + C。

Problem 3: ∫ (eˣ)/(eˣ + 5) dx. Set u = eˣ + 5, du = eˣ dx. Then ∫ (1/u) du = ln|u| + C = ln(eˣ + 5) + C.

题目3：∫ (eˣ)/(eˣ + 5) dx。设 u = eˣ + 5，du = eˣ dx。则 ∫ (1/u) du = ln|u| + C = ln(eˣ + 5) + C。

Problem 4 (definite): ∫₀¹ (x)/√(x² + 1) dx. Let u = x² + 1, du = 2x dx, so x dx = du/2. Limits: when x=0, u=1; when x=1, u=2. Integral = (1/2) ∫₁² u^(−1/2) du = (1/2)[2u^(1/2)]₁² = [√u]₁² = √2 − 1.

题目4（定积分）：∫₀¹ (x)/√(x² + 1) dx。设 u = x² + 1，du = 2x dx，因此 x dx = du/2。上下限：当 x=0，u=1；当 x=1，u=2。积分 = (1/2) ∫₁² u^(−1/2) du = (1/2)[2u^(1/2)]₁² = [√u]₁² = √2 − 1。

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11. Tips for Exam Success | 考试成功小贴士

In an IB exam, always write down your substitution clearly: ‘Let u = …’ and ‘du = … dx’. This shows your thought process and helps secure method marks even if a slip occurs later.

在IB考试中，务必清晰地写下你的换元过程：“设 u = …”以及“du = … dx”。这展示了解题思路，即使后续出现小失误也能获得方法分。

When dealing with definite integrals, immediately find the new u-limits and write them beside the integral. This saves time and reduces the chance of forgetting.

处理定积分时，立即求出新的 u 的积分限并标注在积分旁边。这能节省时间并减少遗忘的可能。

Simplify the integrand in terms of u before integrating. If you are left with any x, re-examine the substitution because the integral must be entirely in u.

积分前先将被积函数用 u 化简。如果化简后仍残留 x，需要重新检查换元，因为积分必须完全用 u 表示。

Check your answer by differentiating. If the derivative returns the original integrand, your answer is correct. This is a powerful verification step in the exam.

通过求导来检查答案。如果求导后得到原始的被积函数，说明答案正确。这是考试中非常有效的验证手段。

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12. Summary and Key Takeaways | 总结与要点回顾

Integration by substitution transforms a difficult integral into a standard one by letting u = inner function. The method requires identifying a function-derivative pair, rewriting the integral, and then handling limits appropriately for definite integrals.

换元积分法通过设 u = 内部函数，将复杂的积分转化为标准积分。该方法要求识别函数-导数对，重写积分，并在定积分中恰当地处理积分限。

Mastering this technique is non-negotiable for IB success because it appears in pure mathematics, as well as in applications such as kinematics and probability density functions. Regular practice with exercises like those in 15G builds the intuition needed to spot substitutions quickly.

掌握这一技巧对于IB的成功至关重要，因为它不仅出现在纯数学中，还应用于运动学和概率密度函数等领域。通过像练习15G这样的题目进行定期练习，可以培养快速识别换元的直觉。

Always remember the key patterns: polynomial powers, trigonometric compositions, f'(x)/f(x), and exponential forms. With these in your toolkit, you will confidently face any substitution problem.

始终牢记关键模式：多项式的幂、三角函数的复合、f'(x)/f(x) 以及指数形式。掌握了这些工具，你就能自信地应对任何换元问题。

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