IB Math: Integration by Substitution — Step-by-Step Methods & Tips | IB数学：换元积分法解题步骤与技巧

📚 IB Math: Integration by Substitution — Step-by-Step Methods & Tips | IB数学：换元积分法解题步骤与技巧

Integration by substitution is one of the most powerful and frequently tested techniques in IB Mathematics. It reverses the chain rule for differentiation and allows us to rewrite a complicated integral into a simpler one.

换元积分法是IB数学中最重要、最常考的积分技巧之一。它是微积分中链式法则的逆运算，通过适当的变量代换，将复杂的积分转化为更容易计算的形式。

1. When to Use Substitution | 何时使用换元积分法

The main signal for using substitution is the presence of a composite function multiplied by its derivative, or an expression that can be rearranged to form such a pattern. In general, if the integrand contains f(g(x)) and also contains g'(x), then setting u = g(x) is a natural choice.

使用换元积分法的主要信号是：被积函数中含有复合函数 f(g(x))，并且同时含有 g'(x) 的因子。更一般地说，如果积分式中出现了 f(g(x))，且表达式中可以整理出 g'(x)，那么设 u = g(x) 通常是自然的思路。

Look for an “inner function” whose derivative is also present, up to a constant factor.
寻找“内层函数”，并且它的导数也出现在被积函数中（允许相差一个常数倍）。
If the integrand is a product of a composite function and the derivative of the inner part, substitution usually works.
如果被积函数是复合函数与内层函数导数的乘积，换元法通常有效。

For example, in ∫ 2x cos(x²) dx, the inner function is x², and its derivative 2x is already present. Thus u = x² is the correct substitution.

例如，对于 ∫ 2x cos(x²) dx，内层函数是 x²，它的导数 2x 已经出现在积分式中，因此应设 u = x²。

2. The Three-Step Core Procedure | 三步核心流程

The substitution method always follows the same logical structure. Master these three steps and you will rarely be lost.

换元积分法的解题过程始终遵循同样的逻辑结构。掌握以下三步，你基本上不会迷失方向。

Step 1: Choose a new variable u = g(x). Usually choose the inner function of a composite expression.

第一步：令新变量 u = g(x)。 通常选择复合表达式中的内层函数。

Step 2: Compute du/dx and rewrite dx in terms of du. From du = g'(x) dx, rearrange to express dx, then replace every x-term and dx in the original integral.

第二步：计算 du/dx，并用 du 表示 dx。 由 du = g'(x) dx 出发，整理出 dx 的表达式，然后将原积分中的所有 x 和 dx 全部替换为 u 和 du。

Step 3: Integrate with respect to u, then substitute back using u = g(x).

第三步：对 u 求积分，最后用 u = g(x) 将结果代回原变量 x。

Consider ∫ x√(x² + 1) dx. Let u = x² + 1, then du = 2x dx, so x dx = ½ du. Hence:

以 ∫ x√(x² + 1) dx 为例。令 u = x² + 1，则 du = 2x dx，所以 x dx = ½ du。因此：

∫ x√(x² + 1) dx = ½ ∫ √u du = ⅓ (x² + 1)^{3/2} + C

3. Choosing the Substitution: u = g(x) | 如何选择 u = g(x)

Choosing the correct substitution is often the hardest part for beginners. There is no single rule that works for every integral, but several reliable strategies exist.

如何选择代换往往是初学者最大的难点。没有一条万能规则适用于所有积分，但有一些非常可靠的策略。

Choose u as the inner function of a composite function, such as u = x³ in ∫ x² sin(x³) dx.
选择复合函数的内层作为 u，例如在 ∫ x² sin(x³) dx 中取 u = x³。
Choose u as the denominator when the numerator is its derivative: ∫ f'(x)/f(x) dx = ln|f(x)| + C.
当分子是分母的导数时，选择分母作为 u：∫ f'(x)/f(x) dx = ln|f(x)| + C。
Choose u as the entire base of a power or root expression, e.g. u = x² + 1 for ∫ 2x(x² + 1)ⁿ dx.
选择幂函数或根式的整个底数作为 u，例如在 ∫ 2x(x² + 1)ⁿ dx 中取 u = x² + 1。
For linear inputs such as ax + b, use u = ax + b first; this includes ∫ sin(ax + b) dx.
对于线性结构 ax + b，先设 u = ax + b；这适用于 ∫ sin(ax + b) dx 等。

When a derivative is not exactly present, you can adjust by a constant factor. For example, in ∫ sin(3x) dx, the derivative of 3x is 3, so dx = du/3.

当导数不是恰好出现时，可以通过常数因子进行调整。例如 ∫ sin(3x) dx 中，3x 的导数是 3，所以 dx = du/3。

Common linear substitutions are summarised below.

常见的线性代换总结如下。

Integral form \| 积分形式	Substitution \| 代换	dx in terms of du
∫ f(ax + b) dx	u = ax + b	dx = du/a
∫ x f(x²) dx	u = x²	dx = du/(2x)
∫ f'(x)/f(x) dx	u = f(x)	dx = du/f'(x)

4. Handling Definite Integrals: Changing Limits | 定积分：换限

For definite integrals, you have two valid options. You may either integrate with substitution and then return to the original variable, or convert the limits to the new variable. The second method is usually shorter and cleaner.

对于定积分，你有两种可行的做法：一种是换元积分后换回原变量，再使用原来的积分上下限；另一种是直接将上下限转换为新变量。第二种方法通常更简洁、更不易出错。

When you choose u = g(x), the new limits are simply u = g(a) and u = g(b), where a and b are the original x-limits.

当你令 u = g(x) 时，新的上下限就是 u = g(a) 和 u = g(b)，其中 a 和 b 是原来的 x 上下限。

Consider the definite integral ∫₀¹ 2x e^{x²} dx. Let u = x². Then du = 2x dx, and the limits change as follows: when x = 0, u = 0; when x = 1, u = 1.

考虑定积分 ∫₀¹ 2x e^{x²} dx。令 u = x²，则 du = 2x dx，上下限变换如下：当 x = 0 时，u = 0；当 x = 1 时，u = 1。

∫₀¹ 2x e^{x²} dx = ∫₀¹ e^u du = e – 1

Notice that once the limits are changed, the original x-limits are no longer used. This avoids the extra step of substituting back.

注意：一旦完成了换限，就不再使用原来的 x 上下限。这种方法可以省去代回原变量的额外步骤。

5. Trigonometric Integrals | 三角函数的积分

Trigonometric integrals often involve a composite trigonometric function such as sinᵏx cos x, tan x, or sec²x. Substitution can reduce them to basic power integrals.

三角函数积分通常包含 sinᵏx cos x、tan x、sec²x 等复合结构。换元法可以把它们化简为基本的幂函数积分。

For ∫ sin³x cos x dx, set u = sin x. Then du = cos x dx, and the integral becomes ∫ u³ du = u⁴/4 + C = ¼ sin⁴x + C.

对于 ∫ sin³x cos x dx，令 u = sin x，则 du = cos x dx，积分变为 ∫ u³ du = u⁴/4 + C = ¼ sin⁴x + C。

The integral of tan x is a classic substitution example. Since tan x = sin x / cos x, let u = cos x, then du = -sin x dx, so:

tan x 的积分是换元法的经典例子。因为 tan x = sin x / cos x，令 u = cos x，则 du = -sin x dx，所以：

∫ tan x dx = -∫ du/u = -ln|cos x| + C = ln|sec x| + C

Similarly, ∫ cot x dx = ln|sin x| + C after substituting u = sin x.

类似地，∫ cot x dx = ln|sin x| + C，只需设 u = sin x。

6. Integrals Involving √(a² − x²) and Related Forms | 含根号的积分

When the integrand contains expressions such as √(a² − x²), √(a² + x²), or √(x² − a²), trigonometric substitution is often the most effective approach.

当被积函数中出现 √(a² − x²)、√(a² + x²) 或 √(x² − a²) 等形式时，三角代换通常是最有效的方法。

Expression \| 表达式	Substitution \| 代换	Simplified form \| 化简结果
√(a² − x²)	x = a sin θ	a cos θ
√(a² + x²)	x = a tan θ	a sec θ
√(x² − a²)	x = a sec θ	a tan θ

For example, to evaluate ∫ √(1 − x²) dx, let x = sin θ. Then dx = cos θ dθ and √(1 − x²) = cos θ, so:

例如，计算 ∫ √(1 − x²) dx 时，令 x = sin θ，则 dx = cos θ dθ，且 √(1 − x²) = cos θ，于是：

∫ √(1 − x²) dx = ∫ cos²θ dθ = ½ θ + ¼ sin 2θ + C

Finally, substitute back using θ = arcsin x and sin 2θ = 2x√(1 − x²):

最后用 θ = arcsin x 和 sin 2θ = 2x√(1 − x²) 代回：

∫ √(1 − x²) dx = ½ arcsin x + ½ x√(1 − x²) + C

7. Exponential and Logarithmic Integrals | 指数与对数积分

Exponential functions combine naturally with substitution because the derivative of e^u is e^u itself. If you can identify u such that du appears in the integrand, the integral often becomes trivial.

指数函数与换元法天然契合，因为 e^u 的导数仍是 e^u。只要找到合适的 u，使 du 出现在被积函数中，积分往往就变得非常简单。

Consider ∫ e^{3x} / (1 + e^{3x}) dx. Let u = 1 + e^{3x}, then du = 3e^{3x} dx, so e^{3x} dx = du/3. Therefore:

考虑 ∫ e^{3x} / (1 + e^{3x}) dx。令 u = 1 + e^{3x}，则 du = 3e^{3x} dx，因此 e^{3x} dx = du/3。于是：

∫ e^{3x} / (1 + e^{3x}) dx = ⅓ ∫ du/u = ⅓ ln|1 + e^{3x}| + C

Another common example is ∫ 1/(x ln x) dx. Let u = ln x, then du = dx/x, so the integral becomes ∫ du/u = ln|ln x| + C.

另一个常见例子是 ∫ 1/(x ln x) dx。令 u = ln x，则 du = dx/x，因此积分变为 ∫ du/u = ln|ln x| + C。

8. Rational Functions and Linear Substitution | 有理函数与线性换元

For rational functions where the numerator is a multiple of the derivative of the denominator, substitution reduces the integral to a simple logarithm.

对于有理函数，如果分子是分母导数乘以某个常数倍，那么换元法可以将积分化为简单的对数形式。

Example: ∫ (2x + 1)/(x² + x + 1) dx. Let u = x² + x + 1, then du = (2x + 1) dx. Therefore the integral is simply ∫ du/u = ln|u| + C, so:

例：∫ (2x + 1)/(x² + x + 1) dx。令 u = x² + x + 1，则 du = (2x + 1) dx。因此积分直接变为 ∫ du/u = ln|u| + C，所以：

∫ (2x + 1)/(x² + x + 1) dx = ln|x² + x + 1| + C

Another important linear substitution is for integrals of the form ∫ 1/(ax + b) dx. Setting u = ax + b gives dx = du/a, hence:

另一个重要的线性换元是形如 ∫ 1/(ax + b) dx 的积分。令 u = ax + b，则 dx = du/a，因此：

∫ 1/(ax + b) dx = (1/a) ln|ax + b| + C

9. Common Mistakes and Pitfalls | 常见错误与陷阱

Substitution is a straightforward technique, but small algebraic errors can ruin an otherwise correct solution. Here are the most common mistakes to avoid.

换元法本身并不复杂，但细微的代数错误可能会毁掉一个本来正确的解答。以下是需要避免的常见错误。

Forgetting to replace dx completely. Every x in the integrand must be written in terms of u and du before integrating.
忘记完全替换 dx。在求积分之前，被积函数中的每一个 x 都必须用 u 和 du 表示。
Choosing u = g(x) but not computing du carefully. Always write du = g'(x) dx before rearranging.
虽然选择了 u = g(x)，但没有仔细计算 du。整理之前一定要先写出 du = g'(x) dx。
Forgetting to change the limits for definite integrals. If you use u-limits, never mix them with original x-limits.
定积分忘记换限。如果使用 u 的上下限，就绝不能和原来的 x 上下限混用。
Omitting the constant of integration C in indefinite integrals.
在不定积分中省略积分常数 C。
Failing to substitute back to the original variable. The final answer must be expressed in x, not in u.
忘记代回原变量。最终答案必须用 x 表示，而不是 u。

10. Practice Problems with Solutions | 练习与解答

Try the following problems before reading the solutions. They cover the most common IB examination patterns.

请先尝试独立完成以下练习，再查看解答。这些题目覆盖了IB考试中最常见的题型。

Problem 1: Evaluate ∫ x³ e^{x⁴} dx.

练习1： 计算 ∫ x³ e^{x⁴} dx。

Solution: Let u = x⁴. Then du = 4x³ dx, so x³ dx = du/4. Thus ∫ x³ e^{x⁴} dx = ¼ e^{x⁴} + C.

解答： 令 u = x⁴，则 du = 4x³ dx，所以 x³ dx = du/4。因此 ∫ x³ e^{x⁴} dx = ¼ e^{x⁴} + C。

Problem 2: Evaluate ∫₀^{π/2} sin x cos x dx using the substitution u = sin x.

练习2： 使用代换 u = sin x 计算 ∫₀^{π/2} sin x cos x dx。

Solution: Let u = sin x. Then du = cos x dx. When x = 0, u = 0; when x = π/2, u = 1. Hence:

解答： 令 u = sin x，则 du = cos x dx。当 x = 0 时，u = 0；当 x = π/2 时，u = 1。因此：

∫₀^{π/2} sin x cos x dx = ∫₀¹ u du = ½

Problem 3: Evaluate ∫ 1/(1 + 4x²) dx.

练习3： 计算 ∫ 1/(1 + 4x²) dx。

Solution: Let u = 2x, so dx = du/2. Then the integral becomes ∫ (1/(1 + u²))(du/2) = ½ arctan u + C = ½ arctan(2x) + C.

解答： 令 u = 2x，则 dx = du/2。积分变为 ∫ (1/(1 + u²))(du/2) = ½ arctan u + C = ½ arctan(2x) + C。

11. Summary and Exam Tips | 总结与考试技巧

Integration by substitution is essentially the chain rule in reverse. The key is identifying the inner function, rewriting dx correctly, and working carefully with limits for definite integrals.

换元积分法本质上是链式法则的逆运算。关键在于识别内层函数、正确改写 dx，并在定积分中仔细处理上下限。

In the IB exam, you should always show the substitution step and state the new variable clearly. For definite integrals, show the new limits alongside the original limits. This not only earns method marks but also helps you avoid mistakes.

在IB考试中，你应该清楚地写出代换步骤和新变量。对于定积分，要将新的上下限与原上下限并列写出。这样既能获得方法分，也能帮助你避免错误。

Finally, check your answer by differentiating it. If the derivative returns to the original integrand, your integration is almost certainly correct. This simple verification takes very little time and is especially valuable for indefinite integrals.

最后，通过对结果求导来检验答案。如果导数恰好等于原被积函数，那么你的积分几乎可以肯定是正确的。这个简单的验证方法耗时极少，对于不定积分尤其有价值。

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IB Math: Integration by Substitution — Step-by-Step Methods & Tips | IB数学：换元积分法解题步骤与技巧