📚 Solving an Integration Problem: pdfjoiner_(4)-157 | 积分难题解析：Edexcel 数学 pdfjoiner_(4)-157
In the journey through A-Level Edexcel Mathematics, mastering integration is a pivotal milestone. Students often encounter a variety of functions combined in products, trig ratios, exponentials, or algebraic expressions. The problem labelled as pdfjoiner_(4)-157 is a typical example drawn from a merged revision file, challenging us to find the indefinite integral of a product of a polynomial and a trigonometric function. This article walks you through a systematic solution, explores alternative methods, highlights common pitfalls, and connects the skills to the Edexcel exam specification.
在A-Level Edexcel 数学的学习中，掌握积分是一个关键里程碑。学生常常会遇到多项式、三角函数、指数函数等组合成的乘积形式。pdfjoiner_(4)-157 这个问题来自一份合并的复习文档，要求我们求解一个多项式与三角函数的乘积的不定积分。本文将系统地呈现解题过程，探索替代方法，指出常见错误，并将相关技巧与Edexcel考试规范紧密联系起来。
1. Understanding the Core Problem | 理解核心问题
The problem asks: Evaluate ∫ (2x + 1) cos x dx. At first glance, we notice it is the product of two distinct types of functions: a linear algebraic term (2x + 1) and the trigonometric function cos x. Direct integration of the product is not possible with basic rules, so we must select a suitable technique. The presence of a polynomial multiplied by cos x immediately suggests integration by parts, a method derived from the product rule for differentiation.
题目要求计算 ∫ (2x + 1) cos x dx。乍一看，我们发现这是两种不同类型的函数的乘积：线性代数项 (2x + 1) 和三角函数 cos x。基本积分规则无法直接处理这种乘积，因此我们必须选择合适的技巧。多项式乘以 cos x 的结构立刻提示我们应该使用分部积分法，这一方法源自微分的乘积法则。
2. Recalling Integration by Parts | 回顾分部积分法
The formula for integration by parts is ∫ u dv = uv − ∫ v du trig, Algebraic, Trigonometric, Exponential), where u is chosen first in that order. Since our integrand contains an algebraic factor (2x + 1) and a trigonometric factor (cos x), LIATE suggests u = 2x + 1 (Algebraic) and dv = cos x dx (Trigonometric). Good choices here simplify the resulting integral ∫ v du.
分部积分公式为 ∫ u dv = uv − ∫ v du。为了高效应用，我们需要决定将被积函数的哪一部分设为 u，哪一部分设为 dv。常用的记忆法是 LIATE（对数、反三角、代数、三角、指数），按照这个顺序优先选择 u。由于被积函数包含代数因子 (2x + 1) 和三角函数因子 (cos x)，LIATE 提示令 u = 2x + 1（代数），dv = cos x dx（三角）。这样的选择能使后续积分 ∫ v du 变得简单。
3. Step-by-Step Solution Using LIATE | 使用 LIATE 的分步求解
Set u = 2x + 1 and dv = cos x dx plifies to (2x + 1) sin x − 2 ∫ sin x dx. Now compute the remaining integral: −2 ∫ sin x dx = −2(−cos x) = 2 cos x. Therefore, the final result is (2x + 1) sin x + 2 cos x + C, where C is the constant of integration. Always remember to add +C for indefinite integrals.
设 u = 2x + 1， dv = cos x dx。对 u 求导：du = 2 dx。对 dv 积分：v = ∫ cos x dx = sin x。代入公式：∫ (2x + 1) cos x dx = (2x + 1)(sin x) − ∫ sin x (2 dx)。这化简为 (2x + 1) sin x − 2 ∫ sin x dx。现在计算剩下的积分：−2 ∫ sin x dx = −2(−cos x) = 2 cos x。因此最终结果为 (2x + 1) sin x + 2 cos x + C，其中 C 是积分常数。对于不定积分，一定不要忘记加上 +C。
4. Visualising the Integration Steps | 可视化积分步骤
It helps to write the work clearly as a table or in a logical flow. Below is a concise tabular representation of the parts chosen and the subsequent manipulations.
将解题过程清晰地列成表格或逻辑流程会很有帮助。下面是所选 u 和 dv 以及后续操作的简明表格表示。
| Step | Expression |
|---|---|
| Choose u | u = 2x + 1 |
| Choose dv | dv = cos x dx |
| Differentiate u | du = 2 dx |
| Integrate dv | v = sin x |
| Apply formula | (2x + 1) sin x − ∫ sin x * 2 dx |
| Simplify | (2x + 1) sin x + 2 cos x + C |
Visualising the process reduces the risk of sign errors and helps in checking the work. Always differentiate the final answer to see if it yields the original integrand.
可视化这个过程可以减少符号错误的风险，并有助于检查。始终对最终答案求导，看看是否得到原来的被积函数。
5. Verifying by Differentiation | 通过求导进行验证
To confirm our result, differentiate (2x + 1) sin x + 2 cos x ive of 2 cos x is −2 sin x. Adding them gives (2x + 1) cos x, which matches the original integrand perfectly. This verification is a powerful habit for Edexcel exams where marks are allocated for checking understanding.
为了确认结果，对 (2x + 1) sin x + 2 cos x 求导。对 (2x + 1) sin x 使用乘积法则：导数 = (2)(sin x) + (2x + 1)(cos x) = 2 sin x + (2x + 1) cos x。然后 2 cos x 的导数是 −2 sin x。将它们相加得到 (2x + 1) cos x，与原始被积函数完全吻合。在Edexcel考试中，这种验证是一个强有力的习惯，能帮助你拿到检查性分数。
6. Alternative Approach: Substitution vs Parts | 替代方法：代换与分部
Could we solve this by substitution instead? (u − 1)/2) * (1/2) du, which introduces a cosine of a linear expression with a fraction. This route becomes significantly messier and does not simplify the trigonometric part. That confirms integration by parts is the optimal method here. Another alternative is to use tabular integration (the 'D.I.' ten used when one part eventually differentiates to zero. Here since 2x + 1 differentiates to a constant, the tabular method works neatly.
我们能否用代换法求解呢？尝试直接 u-代换，比如令 u = 2x + 1，得到 x = (u − 1)/2，dx = du/2。那么积分变为 ∫ u cos((u − 1)/2) * (1/2) du，引入了分数形式的线性余弦，这使得问题变得更加复杂，并没有简化三角函数部分。这证实了分部积分在这里是最佳方法。另一种替代方法是表格积分法（'D.I.'法），它遵循相同的逻辑，但将反复求导和积分的步骤列在表中，常用于其中一个部分最终微分为零的情况。这里 2x + 1 微分后得到常数，表格法同样清晰有效。
7. Handling Similar Edexcel Style Integrals | 处理类似 Edexcel 风格的积分
integration of polynomial × trig or polynomial × exponential functions. For example, ∫ x eˣ ell. However, with logarithmic functions, the logarithm is always chosen as u because its derivative is algebraic, making ∫ v du manageable. Practice swapping u and dv with simple examples to see why wrong choices lead to more complex integrals.
Edexcel A-Level 试卷经常出现多项式乘三角或多项式乘指数函数的积分。例如 ∫ x eˣ dx，∫ (3x² + 1) sin x dx 或 ∫ x² ln x dx。每种情况下，LIATE 规则都指导着 u 的选择。对于指数函数，令代数部分为 u 同样有效。然而，对于对数函数，总是选择对数为 u，因为它的导数是代数形式，能使 ∫ v du 变得易于处理。可以通过简单例子交换 u 和 dv，体会错误选择为什么会导向更复杂的积分。
8. Common Mistakes and Their Remedies | 常见错误与补救方法
- Forgetting the constant +C: In indefinite integration, always write +C. Edexcel often deducts a mark if omitted.
- Sign errors when integrating dv: If dv = cos x dx, then v = sin x, not −sin x. Double-check basic derivatives.
- Incorrectly differentiating u For u = 2x + 1, du = 2 dx, not just 2. The dx must be carried through to the substitution.
- Misapplying the formula: The formula is uv − ∫ v du, not uv + ∫ v du. Keep the minus sign clearly noted.
- Choosing u and dv poorly dx, v = x² + x. Then ∫ v du becomes ∫ (x² + x)(−sin x) dx, which is even more complicated.
Remember that systematic working and re-checking limits (for definite integrals) or differentiation removes these common pitfalls.
- 忘记常数 +C：不定积分中一定要写 +C。Edexcel 常常因遗漏而扣分。
- 积分 dv 时的符号错误：如果 dv = cos x dx，则 v = sin x，而不是 −sin x。重新确认基本导数。
- 错误地对 u 求导：对于 u = 2x + 1，du = 2 dx，而不只是 2。dx 必须贯穿代换过程。
- 误用公式：公式是 uv − ∫ v du，而不是 uv + ∫ v du。要清楚地保留减号。
- 糟糕的 u 和 dv 选择：如果设 u = cos x，会得到 du = −sin x dx，dv = (2x+1)dx，v = x² + x。那么 ∫ v du 变成 ∫ (x² + x)(−sin x) dx，变得更加复杂。
请记住，系统化的解题步骤、重新检查（对定积分是上下限）或求导验证可以消除这些常见陷阱。
9. Connecting to Broader Syllabus and Kinematics | 关联更广泛的教学大纲与运动学
in mechanics when integrating variable acceleration. For instance, if velocity v(t) = (2t + 1) cos t, the displacement is ∫ v(t) dt. The techniques you polish here apply directly to such applied problems. Edexcel often embeds integration by parts within multi-step questions, requiring you to recognise when to use it without being prompted.
分部积分并不是一个孤立的主题；它出现在诸如求曲线下方面积、解微分方程，甚至在力学中当需要对变加速度积分时等情境中。例如，如果速度 v(t) = (2t + 1) cos t，位移就是 ∫ v(t) dt。你在这里磨练的技巧可以直接应用于此类实际问题。Edexcel 经常将分部积分嵌入到多步骤问题中，要求你无需提示就能识别出何时该用它。
10. Practice Problems and Exam Tips | 练习题与备考建议
To solidify your understanding, attempt the following problems without looking at solutions:
1. ∫ x sin x dx
2. ∫ (3x − 2) eˣ dx
3. ∫ (1 + x²) cos 2x dx
4. ∫ ln x dx (Hint: write ln x as 1 * ln x, set u = ln x, dv = 1 dx)
After solving, differentiate your answers as a check. On exam day, manage your time by quickly noting down u, du, dv, v for each integration by parts problem. Ensure your calculator is in the correct mode for definite integrals if allowed. Finally, always read the question carefully—some integrals might be preceded by a differentiation that directly gives the integrand, saving you from doing any integration at all.
为了巩固理解，请尝试在不看答案的情况下解决以下问题：
1. ∫ x sin x dx
2. ∫ (3x − 2) eˣ dx
3. ∫ (1 + x²) cos 2x dx
4. ∫ ln x dx（提示：将 ln x 写成 1 * ln x，设 u = ln x，dv = 1 dx）
解题后，通过求导检查你的答案。在考试当天，合理安排时间，快速写出每个分部积分问题的 u、du、dv、v。如果考试允许使用计算器，确保其在定积分计算时处于正确模式。最后，务必仔细读题——有些积分前面可能有一个直接给出被积函数的微分步骤，这样你甚至无需动手积分。
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