📚 Integration by Substitution | 换元积分法
Integration by substitution is a fundamental technique for evaluating integrals, often considered the reverse of the chain rule for differentiation. In IB Mathematics Analysis & Approaches HL, Exercise 14I typically provides extensive practice on this method, covering both indefinite and definite integrals. Mastering substitution is essential for handling more complex integrals involving composite functions, and it forms a key building block for further topics such as differential equations.
换元积分法是求积分的基本技巧，常被视为链式法则微分的逆运算。在IB数学分析与方法HL课程中，练习14I通常提供了对这种方法的广泛训练，涵盖不定积分和定积分。掌握换元法对于处理涉及复合函数的更复杂积分至关重要，并且它是进一步学习微分方程等主题的关键基石。
1. The Chain Rule in Reverse | 逆链式法则
When you differentiate a composite function, you apply the chain rule: d/dx [f(g(x))] = f'(g(x)) · g' bstitution reverses this process, allowing us to integrate expressions of the form ∫ f'(g(x)) · g' challenge lies in recognising the inner function and its derivative inside the integrand.
当你对复合函数求导时，应用链式法则：d/dx [f(g(x))] = f'(g(x)) · g'(x)。换元积分法通过引入新变量 u = g(x) 逆转这一过程，使我们可以对形如 ∫ f'(g(x)) · g'(x) dx 的表达式进行积分。难点在于识别被积函数中的内层函数及其导数。
∫ f(g(x)) · g'(x) dx = ∫ f(u) du, where u = g(x)
2. The Substitution Method Step by Step | 换元法步骤拆解
k for an inner function whose derivative also appears (up to a constant factor) in the integrand.
步骤1：选择合适的 u = g(x)。寻找一个内层函数，其导数（可能相差常数倍）也出现在被积函数中。
Step 2: Compute du/dx = g'(x) and rewrite this as du = g'(x) dx. This replaces the dx term.
步骤2：计算 du/dx = g'(x) 并改写为 du = g'(x) dx。这一步替换了 dx 项。
Step 3: Substitute all x expressions and dx with u and du, so the integral becomes a function of u only.
步骤3：用 u 和 du 替换所有含 x 的表达式和 dx，使积分变为仅含 u 的函数。
Step 4: Evaluate the simpler integral in terms of u. Use your standard integration rules.
步骤4：计算关于 u 的更简单的积分，运用基本积分公式。
Step 5: For indefinite integrals, rewrite the result back in terms of x by replacing u with g(x). Remember to add the constant of integration C.
步骤5：对于不定积分，将 u 换回 g(x)，用 x 表示结果。记得加上积分常数 C。
3. Choosing the Substitution | 选择换元
= ½ du. The integral becomes ½ ∫ √u du, which is straightforward.
选择复合函数的内层函数作为 u。通常，u = 括号内幂函数、根号内的表达式，或有理函数的分母。对于 ∫ x·√(x²+1) dx，设 u = x²+1，则 du = 2x dx，因此 x dx = ½; du。积分变为 ½ ∫ √u du，计算起来很简单。
∫ x·√(x²+1) dx = ½; ∫ u1/2 du = ½; × (2/3) u3/2 + C = ⅓ (x²+1)3/2 + C
4. Substitution with Definite Integrals | 定积分换元
fter substituting and changing the limits, you evaluate the integral without converting back to x. This saves time and reduces algebraic errors.
对于定积分，必须改变积分上下限以匹配新变量 u。如果 u = g(x)，那么当 x = a 时，u = g(a)；当 x = b 时，u = g(b)。进行代换并改变上下限后，直接计算积分，无需再换回 x。这节省时间并减少代数错误。
Example: ∫&# ²+1)³ dx. Let u = x²+1, du = 2x dx. Limits: x=0 ⇒ u=1; x=1 ⇒ u=2.
示例：∫₀¹ 2x(x²+1)³ dx。令 u = x²+1，du = 2x dx。积分限：x=0 得 u=1；x=1 得 u=2。
∫₀¹ 2x(x²+1)³ dx = ∫₁² u³ du = [¼ u⁴]₁² = 16/4 − 1/4 = 15/4
Notice how the final answer is obtained directly without returning to x. IB exams often test this skill explicitly.
注意最终答案直接求得，无需回到 x。IB考试经常明确考查这一技巧。
5. Trigonometric Substitutions | 三角换元
When the integrand contains √(a²−x²), √(a²+x²), or √(x²−a²), trigonometric substitution often simplifies the integral. The idea is to use a Pythagorean identity to eliminate the square root.
当被积函数含有 √(a²−x²)、√(a²+x²) 或 √(x²−a²) 时，三角换元往往能简化积分。其思路是利用毕达哥拉斯恒等式消去根号。
| Expression | Substitution | Identity Used |
|---|---|---|
| √(a²−x²) | x = a sin &
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