Exercise 15B: Integration by Substitution | 习题15B：换元积分法

📚 Exercise 15B: Integration by Substitution | 习题15B：换元积分法

Integration by substitution is one of the most powerful techniques in calculus, allowing us to transform complex integrals into simpler forms. In Exercise 15B, we practice this method extensively to build fluency. This article walks through the key concepts, step‑by‑step examples, and common pitfalls to help you master this essential skill.

换元积分法是微积分中最强大的技巧之一，能将复杂的积分转化为更简单的形式。在习题15B中，我们通过大量练习来熟练掌握这一方法。本文将梳理核心概念、分步示例以及常见误区，帮助你掌握这项关键技能。

---

1. The Core Idea Behind Substitution | 换元法的核心思想

tegral in terms of u. This simplifies the integrand and makes integration straightforward.

换元法逆用了链式法则。如果被积函数是一个复合函数，且内层函数的导数作为因子出现，我们就可以令 u = g(x)，并将积分用 u 表示。这样做可以简化被积函数，使积分变得直接。

---

2. The Substitution Formula | 换元公式

The general formula for integration by substitution is: ∫ f(g(x))·g'(x) dx = ∫ f(u) du, where u = g(x) and du = g' ter integrating with respect to u, we substitute back to express the result in terms of x.

换元积分法的一般公式为：∫ f(g(x))·g'(x) dx = ∫ f(u) du，其中 u = g(x)，du = g'(x) dx。在对 u 积分后，我们再代回原变量 x 表示结果。

∫ f(g(x))·g'(x) dx = ∫ f(u) du

---

3. How to Choose u | 如何选取 u

Choosing the right u is critical. Typically, u is the inner function of a composition, or a function whose derivative also appears in the integrand. Look for an expression inside parentheses, under a square root, or in the exponent. Often, the derivative of u will be present (up to a constant factor).

正确选取 u 至关重要。通常，u 是复合函数的内层函数，或者其导数也出现在被积函数中。注意寻找括号内、根号下或指数位置上的表达式。通常，u 的导数会在被积函数中出现（可能相差一个常数因子）。

---

4. Example 1: Simple Substitution | 例1：简单换元

Evaluate ∫ 2x·(x² + 1)³ dx.

计算 ∫ 2x·(x² + 1)³ dx。

u/dx = 2x, so du = 2x dx. The integral becomes ∫ u³ du = (1/4)u⁴ + C = (1/4)(x² + 1)⁴ + C.

令 u = x² + 1，则 du/dx = 2x，因此 du = 2x dx。积分变为 ∫ u³ du = (1/4)u⁴ + C = (1/4)(x² + 1)⁴ + C。

∫ 2x (x²+1)³ dx = ¼ (x²+1)⁴ + C

---

5. Example 2: Linear Inside Function | 例2：线性内层函数

Evaluate ∫ cos(3x+2) dx.

计算 ∫ cos(3x+2) dx。

= du/3. The integral becomes ∫ cos(u) · (1/3) du = (1/3) sin(u) + C = (1/3) sin(3x+2) + C.

令 u = 3x+2，则 du = 3 dx → dx = du/3。积分变为 ∫ cos(u) · (1/3) du = (1/3) sin(u) + C = (1/3) sin(3x+2) + C。

∫ cos(3x+2) dx = ⅓ sin(3x+2) + C

---

6. Example 3: Substitution with Polynomial | 例3：多项式换元

Evaluate ∫ x·√x²+4 dx.

计算 ∫ x·√x²+4 dx。

Let u = x²+4, then du = 2x dx, so x dx = du/2. The integral becomes ∫ √u · (1/2) du = (1/2) ∫ u^1/2 du = (1/2)·(2/3)u^3/2 + C = (1/3)(x²+4)^3/2 + C.

令 u = x²+4，则 du = 2x dx，因此 x dx = du/2。积分变为 ∫ √u · (1/2) du = (1/2) ∫ u^1/2 du = (1/2)·(2/3)u^3/2 + C = (1/3)(x²+4)^3/2 + C。

∫ x√x²+4 dx = ⅓ (x²+4)^3/2 + C

---

7. Example 4: Trigonometric Substitution | 例4：三角换元

Evaluate ∫ sin² x cos x dx.

计算 ∫ sin² x cos x dx。

sin x, then du = cos x dx. The integral becomes ∫ u² du = (1/3)u³ + C = (1/3) sin³ x + C.

注意到 sin x 的导数是 cos x，因此令 u = sin x，则 du = cos x dx。积分变为 ∫ u² du = (1/3)u³ + C = (1/3) sin³ x + C。

∫ sin² x cos x dx = ⅓ sin³ x + C

---

广告 / Sponsored

8. Example 5: Exponential Substitution | 例5：指数函数换元

Evaluate ∫ e^3x dx.

计算 ∫ e^3x dx。

Let u = 3x, du = 3 dx → dx = du/3. Then ∫ e^u · (1/3) du = (1/3) e^u + C = (1/3) e^3x + C. This is a simple linear substitution.

令 u = 3x，du = 3 dx → dx = du/3。则 ∫ e^u · (1/3) du = (1/3) e^u + C = (1/3) e^3x + C。这是一个简单的线性换元。

∫ e^3x dx = ⅓ e^3x + C

---

9. Substitution in Definite Integrals | 定积分中的换元

tion. If u = g(x), then when x = a, u = g(a); when x = b, u = g(b). The integral becomes ∫_x=a^x=b f(x) dx = ∫_u=g(a)^u=g(b) f(u) du. This avoids having to substitute back after integrating.

在用换元法计算定积分时，记得要更换积分上下限。如果 u = g(x)，那么当 x = a 时，u = g(a)；当 x = b 时，u = g(b)。积分变为 ∫_x=a^x=b f(x) dx = ∫_u=g(a)^u=g(b) f(u) du。这样就无需在积分后代回原变量。

Example: Evaluate ∫₀¹ 2x (x²+1)³ dx.

例：计算 ∫₀¹ 2x (x²+1)³ dx。

en x=0, u=1; when x=1, u=2. The integral becomes ∫₁² u³ du = [u⁴/4]₁² = 16/4 - 1/4 = 15/4.

令 u = x²+1，du = 2x dx。上下限：x=0 时 u=1；x=1 时 u=2。积分变为 ∫₁² u³ du = [u⁴/4]₁² = 16/4 - 1/4 = 15/4。

∫₀¹ 2x(x²+1)³ dx = 15/4

---

10. Common Mistakes to Avoid | 常见错误与避免方法

• Forgetting to adjust for constant factors: Always check if du matches exactly, or if a constant multiplier is needed.
• Not changing limits in definite integrals: Leads to incorrect results.
• Poor choice of u: If the derivative is not present, the substitution may not simplify the integral.
• Leaving the answer in terms of u: Always substitute back to the original variable (unless it's a definite integral with adjusted limits).

• 忘记调整常数因子：务必检查 du 是否完全匹配，或者是否需要乘以常数。
• 没有更换定积分的上下限：会导致错误结果。
• u 的选择不当：如果导数不在被积函数中，换元可能无法简化积分。
• 答案保留为 u 的函数：必须代回原变量（除非是已更换上下限的定积分）。

---

11. Practice Tips from Exercise 15B | 习题15B的练习建议

some problems that require a clever choice of u, such as ∫ tan x dx = ∫ sin x/cos x dx, letting u = cos x.

习题15B可能包含各种旨在训练模式识别的积分。练习识别内层函数及其导数。完成那些导数相差常数因子的题目，学会通过乘法因子进行调整。同时，尝试一些需要巧妙选取 u 的题目，例如 ∫ tan x dx = ∫ sin x/cos x dx，令 u = cos x。

---

12. Summary | 总结

Integration by substitution is a fundamental tool. By mastering the technique shown in Exercise 15B, you can tackle a wide range of integrals encountered in IB Mathematics. Remember to choose u wisely, handle constants carefully, and always verify your answer by differentiating.

换元积分法是一个基础工具。通过掌握习题15B所展示的方法，你就能应对 IB 数学中遇到的各种积分。记住要明智地选择 u，小心处理常数，并始终通过求导来验证你的答案。

---

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）