IB Math: Integration of Products of Powers of Sine and Cosine — Substitution for Odd Powers | IB数学：正余弦幂乘积积分中奇数次的换元法

📚 IB Math: Integration of Products of Powers of Sine and Cosine — Substitution for Odd Powers | IB数学：正余弦幂乘积积分中奇数次的换元法

Integrals of the form ∫ sin^m x cosⁿ x dx appear frequently in IB Mathematics Analysis and Approaches (AA) HL and occasionally in Analysis and Approaches SL. When at least one of the exponents m or n is a positive odd integer, a clean substitution reveals a simple polynomial integral in disguise. This article explains the method step by step, with full worked examples that match IB mark scheme expectations.

形如 ∫ sin^m x cosⁿ x dx 的积分在 IB 数学分析与方法（AA）HL 中经常出现，在 SL 中也偶有考查。当 m 或 n 中至少有一个是正奇数时，通过一个巧妙的换元，原积分就会变成一个简单的多项式积分。本文将逐步讲解这一方法，并给出与 IB 评分方案完全一致的完整解答示例。

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1. The General Form and Strategy Overview | 一般形式与策略概述

We are concerned with integrals of the form:

∫ sin^m x cosⁿ x dx

where m and n are non-negative integers. The key observation is the derivative relationships: d/dx (sin x) = cos x and d/dx (cos x) = −sin x. If one exponent is odd, we can “peel off” one factor of that function and use it as part of the differential.

我们所关注的积分形式为：

∫ sin^m x cosⁿ x dx

其中 m 和 n 是非负整数。关键观察是导数关系：d/dx (sin x) = cos x，d/dx (cos x) = −sin x。如果其中一个指数是奇数，我们可以“拆出”该函数的一个因子，并把它作为微分的一部分。

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2. Odd Power of Cosine: Substitute u = sin x | 余弦为奇次：令 u = sin x

Suppose n is odd. Write n = 2k + 1, so cosⁿ x = cos^2k x · cos x. Keep one cos x with dx, and replace the remaining even power cos^2k x using the identity cos² x = 1 − sin² x.

Then let u = sin x, du = cos x dx. The integral becomes a polynomial in u:

∫ sin^m x (1 − sin² x)^k cos x dx = ∫ u^m (1 − u²)^k du

假设 n 是奇数。令 n = 2k + 1，则 cosⁿ x = cos^2k x · cos x。保留一个 cos x 与 dx 结合，剩下的偶次幂 cos^2k x 用恒等式 cos² x = 1 − sin² x 替换。

然后令 u = sin x，du = cos x dx。原积分变成关于 u 的多项式积分：

∫ sin^m x (1 − sin² x)^k cos x dx = ∫ u^m (1 − u²)^k du

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3. Worked Example 1: ∫ sin⁴ x cos³ x dx | 示例1：∫ sin⁴ x cos³ x dx

Find ∫ sin⁴ x cos³ x dx.

求 ∫ sin⁴ x cos³ x dx。

Since the power of cosine is odd, reserve one cos x. Write cos³ x = cos² x · cos x = (1 − sin² x) cos x.

因为余弦的幂是奇数，所以保留一个 cos x。将 cos³ x 写成 cos² x · cos x = (1 − sin² x) cos x。

Let u = sin x, du = cos x dx. Then:

令 u = sin x，du = cos x dx。则：

∫ sin⁴ x cos³ x dx = ∫ u⁴ (1 − u²) du = ∫ (u⁴ − u⁶) du

Integrate term by term:

逐项积分：

∫ (u⁴ − u⁶) du = u⁵/5 − u⁷/7 + C

Substitute back u = sin x:

代回 u = sin x：

∫ sin⁴ x cos³ x dx = (sin⁵ x)/5 − (sin⁷ x)/7 + C

This is the final answer. Notice that no trigonometric identities beyond the Pythagorean identity were needed.

这就是最终答案。注意，除了毕达哥拉斯恒等式之外，不需要其他三角恒等式。

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4. Odd Power of Sine: Substitute u = cos x | 正弦为奇次：令 u = cos x

Suppose m is odd. Write m = 2k + 1, so sin^m x = sin^2k x · sin x. Keep one sin x with dx, and replace sin^2k x using sin² x = 1 − cos² x.

Then let u = cos x, du = −sin x dx. The integral becomes:

∫ (1 − cos² x)^k cosⁿ x sin x dx = −∫ (1 − u²)^k uⁿ du

假设 m 是奇数。令 m = 2k + 1，则 sin^m x = sin^2k x · sin x。保留一个 sin x 与 dx 结合，用 sin² x = 1 − cos² x 替换 sin^2k x。

然后令 u = cos x，du = −sin x dx。原积分变为：

∫ (1 − cos² x)^k cosⁿ x sin x dx = −∫ (1 − u²)^k uⁿ du

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5. Worked Example 2: ∫ sin⁵ x cos² x dx | 示例2：∫ sin⁵ x cos² x dx

Find ∫ sin⁵ x cos² x dx.

求 ∫ sin⁵ x cos² x dx。

Since the power of sine is odd, reserve one sin x:

因为正弦的幂是奇数，所以保留一个 sin x：

sin⁵ x cos² x = (sin² x)² cos² x · sin x = (1 − cos² x)² cos² x · sin x

Let u = cos x, du = −sin x dx. Then:

令 u = cos x，du = −sin x dx。则：

∫ sin⁵ x cos² x dx = ∫ (1 − cos² x)² cos² x sin x dx = −∫ (1 − u²)² u² du

Expand the integrand:

展开被积函数：

−∫ (1 − 2u² + u⁴) u² du = −∫ (u² − 2u⁴ + u⁶) du

Integrate:

积分：

− (u³/3 − 2u⁵/5 + u⁷/7) + C = −u³/3 + 2u⁵/5 − u⁷/7 + C

Substitute back u = cos x:

代回 u = cos x：

∫ sin⁵ x cos² x dx = −(cos³ x)/3 + 2(cos⁵ x)/5 − (cos⁷ x)/7 + C

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6. Both Exponents Odd: Choose the Lower Power | 两个指数均为奇数：选择较低次幂

If both m and n are odd, either substitution works. In IB exam questions, you may choose whichever is more convenient. As a general rule, substitute for the function with the lower exponent, because the resulting expansion will be shorter. For example, for ∫ sin³ x cos⁵ x dx, substituting u = sin x gives a quicker expansion than substituting u = cos x.

如果 m 和 n 都是奇数，两种换元都可行。在 IB 考题中，你可以选择更方便的一种。一般原则是：对指数较小的那个函数进行换元，因为这样展开会更简短。例如，对于 ∫ sin³ x cos⁵ x dx，令 u = sin x 比令 u = cos x 展开更快。

Let us compare both approaches for ∫ sin³ x cos⁵ x dx.

我们比较 ∫ sin³ x cos⁵ x dx 的两种方法。

Method 1: u = sin x. Since cos power is odd (5), write cos⁵ x = (1 − sin² x)² cos x. The integral becomes ∫ u³ (1 − u²)² du, which expands to u³ − 2u⁵ + u⁷. This is straightforward.

方法1：u = sin x。因为余弦指数为奇数5，写成 cos⁵ x = (1 − sin² x)² cos x。积分变为 ∫ u³ (1 − u²)² du，展开为 u³ − 2u⁵ + u⁷，非常直接。

Method 2: u = cos x. Since sin power is odd (3), write sin³ x = (1 − cos² x) sin x. The integral becomes −∫ (1 − u²) u⁵ du, which expands to −u⁵ + u⁷. This is even shorter.

方法2：u = cos x。因为正弦指数为奇数3，写成 sin³ x = (1 − cos² x) sin x。积分变为 −∫ (1 − u²) u⁵ du，展开为 −u⁵ + u⁷，甚至更短。

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7. Definite Integrals: Changing the Limits | 定积分：换元时更换上下限

For definite integrals, whenever you substitute, you must also change the limits. For example, consider:

对于定积分，只要进行换元，就必须同时更换上下限。例如，考虑：

∫₀^π/2 sin³ x cos² x dx

Since sin power is odd, let u = cos x. Then du = −sin x dx. When x = 0, u = 1; when x = π/2, u = 0. Also sin² x = 1 − u², so sin³ x = (1 − u²) sin x. The integral becomes:

因为正弦指数为奇数，令 u = cos x。则 du = −sin x dx。当 x = 0 时，u = 1；当 x = π/2 时，u = 0。又 sin² x = 1 − u²，所以 sin³ x = (1 − u²) sin x。积分变为：

∫₁⁰ (1 − u²) u² (−du) = ∫₀¹ (u² − u⁴) du

Now integrate:

现在积分：

[u³/3 − u⁵/5]₀¹ = 1/3 − 1/5 = 2/15

Notice how flipping the limits removes the negative sign. This is a key step that IB mark schemes often check.

注意，交换上下限消掉了负号。这是 IB 评分标准中经常检查的关键步骤。

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8. Why Not for Even Powers? | 为什么偶次幂不能这样做？

If both m and n are even, the substitution method above leaves an extra factor that does not match du. For instance, if m = 2 and n = 2, writing sin² x cos² x = (1 − cos² x) cos² x and trying u = cos x gives an integrand involving sin x, but the leftover is just cos² x; you would need another sin x factor to complete du. Therefore, for even powers we use double-angle identities instead.

如果 m 和 n 都是偶数，上述换元法会留下一个无法与 du 匹配的因子。例如，若 m = 2，n = 2，把 sin² x cos² x 写成 (1 − cos² x) cos² x，然后令 u = cos x，得到的被积函数中确实有 sin x，但剩下的只是 cos² x；你还需要另一个 sin x 因子才能凑成 du。因此，对于偶次幂，我们改用二倍角公式。

For reference, the standard identities used in the even case are:

作为参考，偶次情形下使用的标准恒等式是：

sin² x = (1 − cos 2x)/2, cos² x = (1 + cos 2x)/2, sin x cos x = (sin 2x)/2

These reduce the powers, eventually producing terms that are easy to integrate. The odd-power substitution is generally faster and less error-prone, so IB examiners often design questions with one odd exponent to test it.

这些恒等式可以降低幂次，最终产生易于积分的项。奇次换元通常更快、更不易出错，因此 IB 命题者经常设计含有一个奇指数的题目来考查这种方法。

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9. Common Mistakes and How to Avoid Them | 常见错误及避免方法

• Forgetting the negative sign when substituting u = cos x because d(cos x) = −sin x dx. Always write du = −sin x dx explicitly before integrating.

• 当令 u = cos x 时忘记负号，因为 d(cos x) = −sin x dx。积分前务必明确写出 du = −sin x dx。

• Reserving the wrong factor. For cosⁿ x with n odd, you must keep exactly one cos x for du. Keeping two or zero destroys the substitution.

• 为余弦奇次幂保留错误的因子。对于 n 为奇数的 cosⁿ x，必须恰好保留一个 cos x 来凑 du。保留两个或零个都会破坏换元。

• Forgetting to convert all even powers before substituting. You cannot leave sin² x as “sin² x” in the u-integral; it must become 1 − u² (or similar).

• 换元前忘记转换所有偶次幂。不能把 sin² x 在 u 积分中留作“sin² x”；它必须变成 1 − u²（或其他等价形式）。

• Not adjusting limits for definite integrals. Use the substitution equation u = sin x or u = cos x to compute the new limits, and remember that flipping the limits changes the sign.

• 定积分没有调整上下限。用换元方程 u = sin x 或 u = cos x 计算新的上下限，并记住交换上下限会改变符号。

• Dropping the constant of integration C in indefinite integrals. IB marks deduct for missing C.

• 不定积分漏掉积分常数 C。IB 评分会因漏写 C 而扣分。

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10. IB Exam-Style Questions | IB 风格考题

Let us solve two typical IB-style problems with full working.

下面用完整过程解答两道典型的 IB 风格题目。

Question 1: Find ∫₀^π/4 sin² x cos⁵ x dx.

题目1：求 ∫₀^π/4 sin² x cos⁵ x dx。

Since cos power is odd, let u = sin x, du = cos x dx. Write cos⁵ x = (1 − sin² x)² cos x. Limits: x = 0 ⇒ u = 0; x = π/4 ⇒ u = √2/2. Thus:

因为余弦指数为奇数，令 u = sin x，du = cos x dx。写出 cos⁵ x = (1 − sin² x)² cos x。上下限：x = 0 ⇒ u = 0；x = π/4 ⇒ u = √2/2。因此：

∫₀^π/4 sin² x cos⁵ x dx = ∫₀^√2/2 u² (1 − u²)² du

Expand:

展开：

∫₀^√2/2 (u² − 2u⁴ + u⁶) du = [u³/3 − 2u⁵/5 + u⁷/7]₀^√2/2

Substitute the upper limit:

代入上限：

(√2/2)³/3 − 2(√2/2)⁵/5 + (√2/2)⁷/7

Simplify:

化简：

= (2√2/8)/3 − 2(4√2/32)/5 + (8√2/128)/7 = √2/12 − √2/10 + √2/112

Combining with common denominator 1680:

用公分母 1680 合并：

(140√2 − 168√2 + 15√2)/1680 = −13√2/1680

Wait, this is negative, which seems suspicious because the integrand is non-negative on [0, π/4]. Let us recheck the expansion signs. Actually cos⁵ x = (1 − sin² x)² cos x, so the integrand sin² x (1 − sin² x)² cos x = u² (1 − u²)² = u² − 2u⁴ + u⁶, all positive for u in (0,1). The integral must be positive. Let us compute numerically: u=0.7 gives 0.49 – 0.4802 + 0.117649 = 0.127449, positive. The antiderivative evaluated at upper limit: (√2/2)^3 = √2/4? Wait (√2/2)^3 = (√2)^3/8 = 2√2/8 = √2/4. Then divided by 3 gives √2/12. Correct. (√2/2)^5 = (√2)^5/32 = 4√2/32 = √2/8. Times 2 and divided by 5 gives 2*(√2/8)/5 = √2/20. Wait I wrote 2(4√2/32)/5. 4√2/32 = √2/8, times 2 = √2/4, divided by 5 = √2/20. Not √2/10. I made an error. Let’s recalc carefully. (√2/2)^5 = (√2^5)/(2^5) = (4√2)/32 = √2/8. Then 2*(√2/8) = √2/4. Divide by 5 => √2/20. (√2/2)^7 = (√2^7)/(2^7) = 8√2 / 128 = √2/16. Divide by 7 => √2/112. So positive terms: √2/12 – √2/20 + √2/112. LCM of 12,20,112? 12=2^2*3, 20=2^2*5, 112=2^4*7. LCM = 2^4*3*5*7 = 1680. Numerators: 140, 84, 15 -> 71√2/1680. Positive. Good. Need correct in article.

Thus answer: 71√2/1680. We’ll mention careful arithmetic.

Need include correct final.

Question 2: Find ∫ sin³ x cos⁴ x dx.

Since sin power odd, u=cos x. Then -∫(1-u^2)u^4 du = -∫(u^4 – u^6) du = -u^5/5 + u^7/7 + C. So final = -cos^5 x/5 + cos^7 x/7 + C.

We’ll write.

Need ensure no mistakes. Let’s craft Section 10 with both Qs.

Need be careful with “π/4” entry. Use π.

Also mention “√2/2” etc.

Let’s structure section 10 accordingly.

Need perhaps also include table? Not necessary.

Need final summary and practice.

Let’s build from Section 10 onward.

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10. IB Exam-Style Questions | IB 风格考题

Let us solve two typical IB-style problems with full working.

下面用完整过程解答两道典型的 IB 风格题目。

Question 1: Find ∫₀^π/4 sin² x cos⁵ x dx.

题目1：求 ∫₀^π/4 sin² x cos⁵ x dx。

Since cos power is odd, let u = sin x, du = cos x dx. Write cos⁵ x = (1 − sin² x)² cos x. Limits: x = 0 ⇒ u = 0; x = π/4 ⇒ u = √2/2. Thus:

因为余弦指数为奇数，令 u = sin x，du = cos x dx。写出 cos⁵ x = (1 − sin² x)² cos x。上下限：x = 0 ⇒ u = 0；x = π/4 ⇒ u = √2/2。因此：

∫₀^π/4 sin² x cos⁵ x dx = ∫₀^√2/2 u² (1 − u²)² du

Expand the integrand:

展开被积函数：

u² (1 − 2u² + u⁴) = u² − 2u⁴ + u⁶

Integrate and evaluate:

积分并求值：

[u³/3 − 2u⁵/5 + u⁷/7]₀^√2/2

At u = √2/2:

在 u = √2/2 处：

(√2/2)³/3 − 2(√2/2)⁵/5 + (√2/2)⁷/7

Simplify each term:

化简每一项：

= √2/12 − 2(√2/8)/5 + (√2/16)/7 = √2/12 − √2/20 + √2/112

Using common denominator 1680:

用公分母 1680：

= (140√2 − 84√2 + 15√2)/1680 = 71√2/1680

Thus the exact value is 71√2/1680.

因此精确值为 71√2/1680。

Question 2: Find ∫ sin³ x cos⁴ x dx.

题目2：求 ∫ sin³ x cos⁴ x dx。

Since sin power is odd, let u = cos x, du = −sin x dx. Write sin³ x = (1 − cos² x) sin x. Then:

因为正弦指数为奇数，令 u = cos x，du = −sin x dx。写出 sin³ x = (1 − cos² x) sin x。则：

∫ sin³ x cos⁴ x dx = −∫ (1 − u²) u⁴ du = −∫ (u⁴ − u⁶) du

Integrate:

积分：

= −(u⁵/5 − u⁷/7) + C = −u⁵/5 + u⁷/7 + C

Substitute back u = cos x:

代回 u = cos x：

∫ sin³ x cos⁴ x dx = −(cos⁵ x

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