📚 Mastering Integration by Substitution: Tackling ∫ sin³x cos²x dx | 掌握代入积分法：攻克 ∫ sin³x cos²x dx
This article unpacks a classic A‑ cos²x dx using substitution combined with trigonometric identities. Mastering this technique is essential for the Pure Mathematics 3 and 4 modules, where integration of powers of sine and cosine appears regularly.
本文深入讲解一道经典的 A‑level 积分题，灵感来自一份合并的 Edexcel 历年真题汇编中的第 57 题。我们将全面探索如何利用代换法结合三角恒等式求解 ∫ sin³x cos²x dx。熟练掌握这一技巧对纯数学 3 和 4 模块至关重要，在这两个模块中正弦和余弦幂次的积分频繁出现。
1. Introduction to the Problem | 问题引入
A common challenge in Edexcel A‑ s²x dx fits this pattern perfectly. Students often struggle with choosing the right identity and substitution, so we will break it down step by step.
Edexcel A‑level 数学中一个常见的挑战是：当 sinⁿx 与 cosⁿx 的乘积中至少有一个指数为奇数时如何进行积分。积分 ∫ sin³x cos²x dx 恰好符合这一模式。学生在选择正确的恒等式和代换时经常感到困难，因此我们将逐步拆解。
Given: ∫ sin³x cos²x dx
已知：∫ sin³x cos²x dx
2. Understanding the Integrand | 理解被积函数
(2). A reliable strategy is to reserve one factor of sin x to pair with dx when making a u‑substitution involving cos x, because the derivative of cos x is – . We can then rewrite the remaining even powers of sine in terms of cosine using sin²x = 1 – cos²x.
注意到正弦的幂次为奇数 (3)，余弦的幂次为偶数 (2)。一个可靠的策略是保留一个 sin x 因子，以便在用 cos x 作代换时与 dx 配对，因为 cos x 的导数为 –sin x。然后我们可以利用 sin²x = 1 – cos²x 将剩余的偶次正弦用余弦表示。
sin³x = sin²x · sin x = (1 – cos²x) sin x
sin³x = sin²x · sin x = (1 – cos²x) sin x
3. Choosing the Right Substitution | 选择正确的代换
Let u = cos x. Then du/dx = –sin x, which gives du = –sin x dx, or equivalently sin x dx = – tegral into a polynomial in u, making it immediately integrable with standard power rules.
令 u = cos x。则 du/dx = –sin x，可得 du = –sin x dx，或等价地 sin x dx = –du。这一代换将三角积分转化为 u 的多项式，从而可以直接用标准幂法则积分。
4. Step‑by‑Step Solution: Part 1 – Rewriting the Integrand | 逐步求解：第一部分 – 重写被积函数
express the integrand using the chosen substitution variable. We have sin³x cos²x dx = (1 – cos²x) sin x · cos²x dx. Substituting u = cos x and sin x dx = –du yields the following transformation.
在进行代换之前，先用所选变量表达被积函数。我们有 sin³x cos²x dx = (1 – cos²x) sin x · cos²x dx。代入 u = cos x 与 sin x dx = –du 得到如下变换。
∫ sin³x cos²x dx = ∫ (1 – u²) · u² · (–du) = – ∫ (u² – u⁴) du
∫ sin³x cos²x dx = ∫ (1 – u²) · u² · (–du) = – ∫ (u² – u⁴) du
5. Step‑by‑Step Solution: Part 2 – Simplifying the Integral | 逐步求解：第二部分 – 化简积分
Distribute the minus sign and rewrite the integral in ascending powers of u. This gives a simple polynomial that can be integrated term by term without any trigonometric functions remaining.
分配负号，并按 u 的升幂重写积分。这将得到一个简单的多项式，可以逐项积分，不再包含任何三角函数。
– ∫ (u² – u⁴) du = ∫ (u⁴ – u²) du
– ∫ (u² – u⁴) du = ∫ (u⁴ – u²) du
6. Integrating the Polynomial in u | 对 u 的多项式积分
Apply the power rule for integration: ∫ uⁿ du = uⁿ⁺¹/(n+1) + C for any real n ≠ –1. Integrate each term separately.
应用积分幂法则：对于任意实数 n ≠ –1，有 ∫ uⁿ du = uⁿ⁺¹/(n+1) + C。分别对每一项积分。
∫ u⁴ du = (1/5) u⁵ + C
∫ u⁴ du = (1/5) u⁵ + C
∫ u² du = (1/3) u³ + C
∫ u² du = (1/3) u³ + C
Therefore, the indefinite integral in u is (1/5) u⁵ – (1/3) u³ + C.
因此，关于 u 的不定积分为 (1/5) u⁵ – (1/3) u³ + C。
7. Back‑Substitution to x | 回代到 x
Replace u with cos x to express the final answer in terms of the original variable. The constant of integration C remains unchanged.
将 u 替换回 cos x，用原变量表达最终答案。积分常数 C 保持不变。
∫ sin³x cos²x dx = (1/5) cos⁵x – (1/3) cos³x + C
∫ sin³x cos²x dx = (1/5) cos⁵x – (1/3) cos³x + C
This is the complete general antiderivative. It can be factorised as (1/15) cos³x (3 cos²x – 5) + C, but the expanded form is usually accepted in exams.
这就是完整的通解不定积分。它可以因式分解为 (1/15) cos³x (3 cos²x – 5) + C，但在考试中展开形式通常也被接受。
8. Verification by Differentiation | 通过求导验证
iderivative with respect to x using the chain rule. The derivative of cos⁵x is 5 cos⁴x · (–sin x) = –5 sin x cos⁴x. The derivative of cos³x is 3 cos²x · (–sin x) = –3 sin x cos²x. Multiply by the coefficients and simplify.
为验证结果，利用链式法则对求得的原函数关于 x 求导。cos⁵x 的导数为 5 cos⁴x · (–sin x) = –5 sin x cos⁴x。cos³x 的导数为 3 cos²x · (–sin x) = –3 sin x cos²x。乘上系数并化简。
d/dx[ (1/5) cos⁵x – (1/3) cos³x ] = –sin x cos⁴x + sin x cos²x
d/dx[ (1/5) cos⁵x – (1/3) cos³x ] = –sin x cos⁴x + sin x cos²x
Factor out sin x cos²x: sin x cos²x (1 – s²x · sin²x = sin³x cos²x, which matches the original integrand. The solution is verified.
提取公因子 sin x cos²x：sin x cos²x (1 – cos²x) = sin x cos²x · sin²x = sin³x cos²x，与原被积函数一致。解得以验证。
9. Common Pitfalls and Tips | 常见错误与提示
Many students forget the negative sign from du = – stitution when both powers are even or when the differential does not appear; in such cases double‑angle identities are more suitable.
许多学生忘记 du = –sin x dx 中的负号，导致最终答案的符号错误。一定要明确写出代换方程。另一个常见错误是当两个幂次均为偶数或微分不出现时强行代换；此时应使用倍角公式更为合适。
- If the power of sine is odd, reserve one sin x for du and convert the rest using sin²x = 1 – cos²x.
- 如果正弦的幂次为奇数，保留一个 sin x 用于 du，其余用 sin²x = 1 – cos²x 转换。
- If the power of cosine is odd, reserve one cos x for du and use cos²x = 1 – sin²x.
- 如果余弦的幂次为奇数，保留一个 cos x 用于 du，并使用 cos²x = 1 – sin²x。
- When both powers are even, reduce the integrand using half‑angle or double‑angle identities before integrating.
- 当两个幂次均为偶数时，先利用半角或倍角公式化简被积函数再积分。
10. Extension: What if the Powers Change? | 扩展：如果幂次变化呢？
is odd. Reserve one cos x factor and let u = sin x. Then du = cos x dx. Rewrite cos²x as 1 – sin²x and proceed. The integral becomes ∫ sin²x (1 – sin²x) cos x dx = ∫ u² (1 – u²) du, resulting in (1/3) sin³x – (1/5) sin⁵x + C.
考虑一个相关积分：∫ sin²x cos³x dx。这里余弦的幂次为奇数。保留一个 cos x 因子，令 u = sin x。则 du = cos x dx。将 cos²x 改写为 1 – sin²x 并继续。积分化为 ∫ sin²x (1 – sin²x) cos x dx = ∫ u² (1 – u²) du，结果为 (1/3) sin³x – (1/5) sin⁵x + C。
Modifying the exponents teaches you to recognise symmetry and adapt the method. The core idea remains to isolate the odd‑powered function's derivative and convert the rest.
调整指数可以训练你识别对称性并灵活运用方法。核心思想始终是分离出奇次函数的导数并将余下部分转换。
11. Exam‑Style Practice Question | 考试风格练习题
Try this typical Edexcel question that combines integration with a definite upper limit.
请尝试以下典型的 Edexcel 风格题目，该题将积分与定上限结合。
Evaluate ∫₀π/2 sin³x cos²x dx.
计算 ∫₀π/2 sin³x cos²x dx。
rivative we found and apply the limits. cos(π/2) = 0, cos(0) = 1. The value is [ (1/5)(0)⁵ – (1/3)(0)³ ] – [ (1/5)(1)⁵ – (1/3)(1)³ ] = –(1/5 – 1/3) = –(3/15 – 5/15) = 2/15. Always check the sign carefully when substituting limits.
使用我们求得的原函数并代入上下限。cos(π/2) = 0，cos(0) = 1。值为 [ (1/5)(0)⁵ – (1/3)(0)³ ] – [ (1/5)(1)⁵ – (1/3)(1)³ ] = –(1/5 – 1/3) = –(3/15 – 5/15) = 2/15。代入上下限时务必仔细检查符号。
12. Conclusion | 结论
Integrating products like sin³x cos²x dx showcases the power of substitution when odd powers are present. By isolating a factor of sin x and converting the rest via the Pythagorean identity, the integral reduces to a simple polynomial. Mastering this technique will serve you well in Edexcel Pure Mathematics examinations and beyond.
积分形如 sin³x cos²x dx 的乘积展示了当存在奇次幂时代换的强大威力。通过分离出一个 sin x 因子并利用勾股恒等式转换其余部分，积分化归为简单的多项式。掌握这一技巧将在 Edexcel 纯数学考试及更高层次学习中助你一臂之力。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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