📚 Mastering Integration by Substitution: Exercise 25E | 掌握换元积分法：练习 25E
Integration by substitution is one of the most powerful techniques in calculus, often introduced as the reverse of the chain rule for differentiation. In IB Mathematics, Exercise 25E typically builds on basic integration skills and challenges you to recognise when a change of variable can transform a complicated integral into a standard form. Mastering this method not only boosts your confidence but also deepens your understanding of how functions behave under transformation.
换元积分法是微积分中最强大的技巧之一，通常作为微分链式法则的逆运算来引入。在 IB 数学中，练习 25E 通常在基础积分技能之上，挑战你识别何时可以通过变量代换将复杂的积分转化为标准形式。掌握这一方法不仅能增强你的信心，还能加深你对函数在变换下行为的理解。
1. Understanding the Chain Rule in Reverse | 逆向链式法则的理解
When we differentiate a composite function such as sin(x²), the chain rule gives 2x·cos(x²). Integration by substitution reverses this process: if we spot an integrand containing a function and its derivative, we can 'undo' the chain rule. The key idea is to let a new variable u represent the inner function, so that the derivative du/dx helps absorb the extra factor.
当我们对复合函数比如 sin(x²) 求导时，链式法则给出 2x·cos(x²)。换元积分法逆转了这一过程：如果被积函数包含一个函数及其导数，我们就可以“撤销”链式法则。核心思路是让一个新的变量 u 代表内层函数，从而利用导数 du/dx 来吸收额外的因子。
du, which is simply sin(u) + C. Substituting back gives sin(x²) + C, exactly the antiderivative we expect.
例如，考虑 ∫ 2x·cos(x²) dx。设 u = x²，则 du/dx = 2x，即 du = 2x dx。积分就变成了 ∫ cos(u) du，结果就是 sin(u) + C。回代后得到 sin(x²) + C，正是我们预期的原函数。
2. The Substitution Method: Changing Variables | 换元方法：变量代换
ormal method involves three steps: choose a suitable substitution u = g(x), compute du = g' l variable x. This technique works beautifully when the integrand contains a composition g'(x)·f(g(x)).
正式方法包含三个步骤：选择合适的代换 u = g(x)，计算 du = g'(x) dx，并将积分完全用 u 重写。积分完成后，必须转换回原变量 x。当被积函数包含复合形式 g'(x)·f(g(x)) 时，这一技巧非常有效。
Sometimes the integrand does not precisely match g'(x)·f(g(x)), but by multiplying or dividing by a constant, we can force it into that pattern. This is perfectly legitimate because constants can be taken out of the integral.
有时被积函数并不精确地匹配 g'(x)·f(g(x))，但通过乘以或除以一个常数，我们可以强行变成那种形式。这是完全合法的，因为常数可以移到积分号外面。
3. Choosing the Right Substitution u | 选择正确的代换变量 u
Experienced integrators often look for a function inside another function—such as the argument of a trigonometric function, an exponent, or the base of a root. The derivative of this inner function should appear (perhaps multiplied by a constant) elsewhere in the integrand. If so, that inner function is a strong candidate for u.
有经验的积分者通常会寻找一个函数内部的函数——比如三角函数的参数、指数函数的指数、或是根号的底数。该内层函数的导数应该（可能乘以一个常数）出现在被积函数的其他地方。如果是这样，那个内层函数就是 u 的有力候选。
ln x for logarithmic cases. In Exercise 25E you will practise spotting these patterns quickly.
常见的选择包括线性代换 u = ax + b，二次表达式 u = x² + c，三角积分中的 u = sin x 或 u = cos x，以及对数情形下的 u = ln x。在练习 25E 中，你将练习快速识别这些模式。
4. Step-by-Step Procedure for Indefinite Integrals | 不定积分的逐步求解步骤
Let's outline a clear procedure to follow whenever you attempt a substitution for an indefinite integral:
让我们概括一个清晰的步骤，每当尝试不定积分代换时都可以遵循：
Step 1: Identify the inner function u = g(x) and write it down. Step 2: Differentiate to find du/dx and express du as g'(x) dx. Step 3: Rearrange the original integral to isolate the g'(x) dx part and replace with du. Step 4: Perform the integration with respect to u. Step 5: Substitute back g(x) for u and add the constant of integration +C.
第一步：识别内层函数 u = g(x) 并写下。第二步：求导得出 du/dx，并将 du 表示为 g'(x) dx。第三步：重新整理原积分，分离出 g'(x) dx 部分并用 du 替换。第四步：对 u 进行积分。第五步：将 u 换回 g(x) 并加上积分常数 +C。
This systematic approach prevents algebraic slips and ensures you never forget to convert the differential properly. Practise writing every step, especially in exams where method marks are awarded.
这种系统的方法可以防止代数失误，并确保你不会忘记正确转换微分。要练习写下每一步，尤其是在考试中，方法步骤是有分的。
5. Worked Example 1: Polynomial Substitution | 例题1：多项式代换
Evaluate ∫ 3x²·(x³ + 1)⁴ dx.
计算 ∫ 3x²·(x³ + 1)⁴ dx。
+ C. Substituting back yields (x³ + 1)⁵/5 + C. This is a classic polynomial substitution.
注意到 x³ + 1 的导数是 3x²，正好作为因子出现在幂次前面。设 u = x³ + 1，则 du = 3x² dx。积分变为 ∫ u⁴ du = (u⁵)/5 + C。代回得到 (x³ + 1)⁵/5 + C。这是一个经典的多项式代换。
stant: ∫ 6x²·(x³+1)⁴ dx = 2 ∫ 3x²·(x³+1)⁴ dx = 2·(x³+1)⁵/5 + C. The method remains robust.
如果因子是 6x² 而不是 3x²，我们只需用常数进行调整：∫ 6x²·(x³+1)⁴ dx = 2 ∫ 3x²·(x³+1)⁴ dx = 2·(x³+1)⁵/5 + C。该方法仍然稳健。
6. Worked Example 2: Trigonometric Substitution | 例题2：三角代换
Find ∫ sin³x·cos x dx.
求 ∫ sin³x·cos x dx。
n⁴x /4 + C. This works because cos x dx is exactly du, and the remaining sin³x becomes u³.
令 u = sin x，则 du = cos x dx。积分变为 ∫ u³ du = u⁴/4 + C = sin⁴x /4 + C。之所以有效，是因为 cos x dx 正好是 du，余下的 sin³x 就变成了 u³。
What if the integrand were sin³x·cos³x? u = sin x, but we would need to handle the extra cos²x factor using the identity cos²x = 1 – polynomial in u. Exercise 25E often mixes such trigonometric identities with substitution.
如果被积函数是 sin³x·cos³x 呢？我们仍然可以设 u = sin x，但需要利用恒等式 cos²x = 1 – sin²x 来处理多余的 cos²x 因子，从而将积分转化为 u 的多项式。练习 25E 经常将这类三角恒等式与代换结合在一起。
7. Dealing with Definite Integrals: Changing Limits | 处理定积分：更改积分限
When applying substitution to a definite integral, you have two choices: either substitute back to x and then use the original x-limits, or change the limits of integration to match the new variable u. The second method is usually cleaner and avoids the extra step of converting back.
当对定积分应用代换时，有两种选择：要么代换回 x 再使用原始的 x 积分限，要么将积分限更改为与新变量 u 匹配。第二种方法通常更简洁，避免了转换回去的额外步骤。
ecomes g(a), and the upper limit b becomes g(b). Then evaluate the definite integral in u without ever returning to x.
要更改积分限，只需将原始的 x 值代入代换式 u = g(x)。下限 a 变为 g(a)，上限 b 变为 g(b)。然后直接在 u 中计算定积分，无需回到 x。
8. Worked Example 3: Definite Integral with Substitution | 例题3：带代换的定积分
Evaluate ∫₀¹ 2x·√x²+1 dx using substitution.
使用代换计算 ∫₀¹ 2x·√x²+1 dx。
= x² + 1. Then du = 2x dx. When x = 0, u = 1; when x = 1, u = 2. The integral becomes ∫₁² √u du = ∫₁² u1/2 du. Now integrate: [ (2/3)u3/2 ]₁² = (2/3)[23/2 – 13/2] = (2/3)(2√2 – 1). This definite integral illustrates the elegance of changing limits.
设 u = x² + 1，则 du = 2x dx。当 x = 0 时，u = 1；当 x = 1 时，u = 2。积分变为 ∫₁² √u du = ∫₁² u1/2 du。现在积分：[(2/3)u3/2]₁² = (2/3)[23/2 – 13/2] = (2/3)(2√2 – 1)。这个定积分体现了更改积分限的优雅。
9. Recognizing Patterns: Linear Functions Inside Functions | 识别模式：函数内的线性函数
du/a). This is particularly useful for exponential, trigonometric, and rational functions.
有时内层函数是线性的，比如 u = 2x+3。这种代换很简单，因为 du = 2 dx，即 dx = du/2。任何形如 ∫ f(ax+b) dx 的积分都可以通过代换 u = ax+b 简化，得到 ∫ f(u) (du/a)。这对于指数函数、三角函数和有理函数尤其有用。
For example, ∫ e5x dx becomes (1/5)∫ eu du = (1/5)eu + C = (1/5)e5x + C. Exercise 25E includes numerous examples that reinforce this linear adjustment, making it second nature.
例如，∫ e5x dx 变成 (1/5)∫ eu du = (1/5)eu + C = (1/5)e5x + C。练习 25E 包含大量示例来强化这种线性调整，使之成为本能。
10. Advanced Tip: Inverse Substitution and Trigonometric Identities | 高级技巧：逆代换与三角恒等式
u, known as inverse substitution. A classic example is x = sin θ for integrands involving √1–x² ies using identities. The key is to also change dx = cos θ dθ and change limits if needed.
有时用 u 表示 x 的逆代换更方便，例如对于包含 √1–x² 的被积函数，令 x = sin θ。这将积分转化为可利用恒等式化简的三角形式。关键是要同时将 dx 变为 cos θ dθ，并根据需要更改积分限。
For instance, ∫ dx/√1–x² inverse substitution appears later in some syllabi, Exercise 25E may offer a first taste.
例如，∫₀^(1/2) dx/√1–x²：令 x = sin θ，则 dx = cos θ dθ，积分限变为 0 到 π/6，被积函数变为 cos θ / cos θ = 1，得到 ∫₀^(π/6) dθ = π/6。尽管逆代换在部分大纲中稍后出现，但练习 25E 可能让你初步体验。
11. Common Mistakes to Avoid | 常见错误避免
tants: if du = 2x dx but you have 3x dx, you must introduce a factor of 3/2 to compensate.
最常犯的错误之一是忘记将微分 dx 正确地转换为 du。学生有时会写成 ∫ f(u) dx 而不是 ∫ f(u) du。务必检查积分是否完全以 u 表示，包括微分。另一个错误是没有调整常数：如果 du = 2x dx 但你只有 3x dx，你必须引入一个 3/2 的因子来补偿。
With definite integrals, students often mix up the limits by substituting incorrectly or forgetting to change them altogether. Using u-limits directly avoids errors, but only if you compute them correctly. Lastly, never leave the answer in terms of u for an indefinite integral; always substitute back to the original variable.
在定积分中，学生经常因为代换错误或完全忘记更改积分限而混淆它们。直接使用 u 限可以避免错误，但前提是你计算正确。最后，对于不定积分，千万不要把答案留在 u 中；一定要代回原变量。
12. Practice Summary and Key Takeaways | 练习总结与要点提炼
Integration by substitution is a skill that improves with targeted practice. Exercise 25E provides a variety of integrals carefully designed to reinforce pattern recognition, algebraic manipulation, and the disciplined use of the method. Remember these key points: look for a function and its derivative, choose u wisely, adapt constants if necessary, and change limits for definite integrals.
换元积分法是一项通过有针对性的练习可以提升的技能。练习 25E 提供了各种精心设计的积分题，以强化模式识别、代数运算和有条不紊的方法运用。记住这些要点：寻找一个函数及其导数，明智地选择 u，必要时调整常数，并对定积分更改积分限。
As you work through the exercises, try to verbalise the reason behind each substitution. This builds a deeper intuition that will serve you well in more advanced calculus topics, such as integration by parts and differential equations. Consistent practice will make substitution feel as natural as applying the power rule.
在做练习时，试着说出每次代换背后的理由。这会建立起更深层次的直觉，在更高级的微积分主题中，如分部积分法和微分方程，都将使你受益匪浅。持续的练习将使代换如同运用幂法则一样自然。
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