📚 Integration Using Trigonometric Identities (Edexcel A-Level Pure Maths) | 利用三角恒等式进行积分 (Edexcel A-Level 纯数学)
In Edexcel A-Level Pure Mathematics, integration of trigonometric functions often extends beyond the elementary antiderivatives you learn early in the course. Expressions like sin²x, cos³x, or tan x sec³x cannot be integrated directly with the standard table of integrals. Instead, you need to deploy a range of trigonometric identities to rewrite the integrand into a manageable form. This article systematically explores the strategies, key identities, and worked examples required to master trigonometric integration for the Edexcel specification.
在 Edexcel A-Level 纯数学中，三角函数的积分常常超出课程初期所学习的基本原函数。诸如 sin²x、cos³x 或 tan x sec³x 的表达式无法直接利用标准积分表进行积分。相反，你需要运用一系列三角恒等式将被积函数改写为易于处理的形式。本文系统地探讨了掌握 Edexcel 考试大纲所要求的三角积分法所需的策略、关键恒等式以及典型例题。
1. Standard Integrals and the Need for Identities | 标准积分与使用恒等式的必要性
osec x cot x dx = −cosec x + C, and ∫ cosec²x dx = −cot x + C. These results alone, however, are insufficient when powers, products, or composite angles appear.
你应掌握的基本三角积分包括：∫ sin x dx = −cos x + C、∫ cos x dx = sin x + C、∫ sec²x dx = tan x + C、∫ sec x tan x dx = sec x + C、∫ cosec x cot x dx = −cosec x + C 以及 ∫ cosec²x dx = −cot x + C。然而，当出现幂次、乘积或复合角时，仅靠这些结果是不够的。
ms the integrand into a sum of simple terms. Thus, trigonometric identities become indispensable tools that bridge the gap between a seemingly impossible integral and an elementary antiderivative.
以 ∫ sin²x dx 为例。没有哪条标准积分直接对应 sin²x。此时，恒等式 sin²x = ½(1 − cos 2x) 将被积函数转化为简单项之和。因此，三角恒等式成为连接看似不可能的积分与初等原函数之间不可或缺的桥梁。
2. Key Trigonometric Identities to Recall | 需要牢记的关键三角恒等式
are foundational: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ. These allow you to express one function in terms of another, which is vital for substitution methods.
在处理积分之前，你必须牢固掌握在 Edexcel 题型中反复出现的恒等式。毕达哥拉斯恒等式是基础：sin²θ + cos²θ = 1、1 + tan²θ = sec²θ 以及 1 + cot²θ = cosec²θ。这些可以使你将一个函数用另一个表示，这对换元法至关重要。
le identities cos²θ = ½(1 + cos 2θ) and sin²θ = ½(1 − cos 2θ), which are used to reduce even powers.
双角公式同样重要：sin 2θ = 2 sin θ cos θ，cos 2θ = cos²θ − sin²θ = 2 cos²θ − 1 = 1 − 2 sin²θ。将这些公式变形可得半角恒等式 cos²θ = ½(1 + cos 2θ) 与 sin²θ = ½(1 − cos 2θ)，它们用于降低偶次幂。
A + B) − sin(A − B)], cos A cos B = ½[cos(A + B) + cos(A − B)], sin A sin B = ½[cos(A − B) − cos(A + B)].
积化和差恒等式将乘积转化为和，从而实现逐项积分：sin A cos B = ½[sin(A + B) + sin(A − B)]，cos A sin B = ½[sin(A + B) − sin(A − B)]，cos A cos B = ½[cos(A + B) + cos(A − B)]，sin A sin B = ½[cos(A − B) − cos(A + B)]。
Key identities at a glance:
| Identity | Form |
|---|---|
| sin²θ + cos²θ | 1 |
| 1 + tan²θ | sec²θ |
| sin 2θ | 2 sin θ cos θ |
| cos 2θ | 2 cos²θ − 1 = 1 − 2 sin²θ |
| cos²θ | ½(1 + cos 2θ) |
| sin²θ | ½(1 − cos 2θ) |
3. Integrating sinᵐx cosⁿx – Odd Powers | 积分 sinᵐx cosⁿx – 奇次幂
When integrating a product of the form sinᵐx cosⁿx, if at least one of the exponents m or n is an odd positive integer, the strategy is to save one copy of the trigonometric function that has the odd power for the differential, and convert the rest using the Pythagorean identity.
当对 sinᵐx cosⁿx 形式的乘积进行积分时，如果指数 m 或 n 中至少有一个是正的奇数，其策略是为微分保留一个具有奇次幂的三角函数，并利用毕达哥拉斯恒等式转化其余部分。
sin x. Leave the lone sin x with dx to form the differential of −cos x. Express the remaining even power sinᵐ⁻¹x as (sin²x)(m−1)/2 and replace sin²x by 1 − cos²x. Then use the substitution u = cos x, du = − sin x dx.
情况1：m 为奇数。将 sinᵐx 写成 sinᵐ⁻¹x sin x。让单独的 sin x 与 dx 结合构成 −cos x 的微分。将剩余的偶次幂 sinᵐ⁻¹x 表示为 (sin²x)(m−1)/2，并用 1 − cos²x 替换 sin²x。然后使用代换 u = cos x，du = − sin x dx。
Case 2: n is odd. Save a factor cos x and convert cosⁿ⁻¹x = (cos²x)(n−1)/2 to (1 − sin²x)(n−1)/2. Substitute u = sin x, du = cos x dx.
情况2：n 为奇数。保留一个 cos x 因子，并将 cosⁿ⁻¹x = (cos²x)(n−1)/2 转化为 (1 − sin²x)(n−1)/2。代换 u = sin x，du = cos x dx。
becomes −∫ (1 − u²) u² du = −∫ (u² − u⁴) du = −(u³/3 − u⁵/5) + C = (u⁵/5) − (u³/3) + C = (cos⁵x)/5 − (cos³x)/3 + C.
例题：计算 ∫ sin³x cos²x dx。这里 m = 3 为奇数。改写为 ∫ sin²x sin x cos²x dx = ∫ (1 − cos²x) cos²x sin x dx。令 u = cos x，du = − sin x dx。积分变为 u = −∫ (u² − u⁴) du = −(u³/3 − u⁵/5) + C = (u⁵/5) − (u³/3) + C = (cos⁵x)/5 − (cos³x)/3 + C。
4. Integrating Even Powers Using Double-Angle Formulas | 利用双角公式积分偶次幂
d sin²x = ½(1 − cos 2x) to reduce the powers until each term is integrable. This often produces polynomials in cos 2x, which may need further reduction if high powers remain.
当 sinᵐx cosⁿx 中的 m 与 n 均为偶数时，奇次幂方法不再适用。此时需反复应用半角恒等式 cos²x = ½(1 + cos 2x) 和 sin²x = ½(1 − cos 2x) 降次，直到各项均可积分。这通常会生成关于 cos 2x 的多项式，若仍存在高次幂则需进一步降次。
2x). Then replace cos²2x with ½(1 + cos 4x) to obtain ¼[1 + 2 cos 2x + ½(1 + cos 4x)] = ¼ + ½ cos 2x + ⅛ + ⅛ cos 4x = ⅜ + ½ cos 2x + ⅛ cos 4x. Integration then yields ⅜ x + ¼ sin 2x + ⅛(¼ sin 4x) + C = ⅜ x + ¼ sin 2x + ¹⁄₃₂ sin 4x + C.
例如，∫ sin²x dx = ∫ ½(1 − cos 2x) dx = ½x − ¼ sin 2x + C。更复杂的情形如 ∫ cos⁴x dx：将 cos⁴x = (cos²x)² = [½(1 + cos 2x)]² = ¼(1 + 2 cos 2x + cos²2x)。然后将 cos²2x 替换为 ½(1 + cos 4x)，得到 ¼[1 + 2 cos 2x + ½(1 + cos 4x)] = ¼ + ½ cos 2x + ⅛ + ⅛ cos 4x = ⅜ + ½ cos 2x + ⅛ cos 4x。积分后可得 ⅜ x + ¼ sin 2x + ⅛(¼ sin 4x) + C = ⅜ x + ¼ sin 2x + ¹⁄₃₂ sin 4x + C。
This approach works for any even powers, but patience is needed to expand and integrate term by term. Edexcel exam questions often stop at the fourth power.
该方法适用于所有偶次幂，但需要耐心地展开并逐项积分。Edexcel 考题通常最高到四次幂。
5. Integrating tanᵐx secⁿx – Strategy | 积分 tanᵐx secⁿx 的策略
s even, save a factor sec²x to form the differential of tan x. Write secⁿx = secⁿ⁻²x sec²x
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