IB Math: Applications and Techniques of Integration by Substitution | IB数学：换元积分法的应用与技巧

📚 IB Math: Applications and Techniques of Integration by Substitution | IB数学：换元积分法的应用与技巧

Integration by substitution is one of the most powerful techniques in calculus, often described as the reverse of the chain rule. It allows us to simplify complicated integrals by changing the variable of integration.

换元积分法是微积分中最强大的技巧之一，通常被称为链式法则的逆运算。它通过改变积分变量，帮助我们简化复杂的积分。

1. Understanding the Substitution Rule | 理解换元法则

The substitution rule is based on the idea that a well-chosen new variable can transform a difficult integral into a simpler one. If we set u = g(x), then du = g'(x) dx, and the integral becomes an integral in terms of u.

换元法则的基本思想是：选择一个合适的新变量，可以把一个困难的积分转化为更简单的积分。如果令 u = g(x)，则 du = g'(x) dx，原积分就变成关于 u 的积分。

∫ f(g(x)) g'(x) dx = ∫ f(u) du

This formula shows that the derivative g'(x) must appear, possibly up to a constant factor, for the substitution to work cleanly.

这个公式表明，被积表达式中必须出现 g'(x) 的因子，最多相差一个常数倍数，换元才能顺利进行。

2. Step-by-Step Procedure | 分步解题流程

To apply integration by substitution successfully, follow a systematic procedure. This reduces the chance of algebraic errors and helps you recognise patterns quickly.

要成功运用换元积分法，需要遵循一套系统化的步骤。这样可以减少代数错误，并帮助你快速识别题型模式。

Choose a suitable substitution u = g(x) and compute du = g'(x) dx.
选择合适的代换 u = g(x)，并计算 du = g'(x) dx。
Rewrite the entire integral in terms of u and du, eliminating all x and dx.
将整个积分改写成关于 u 和 du 的形式，消去所有 x 和 dx。
Integrate with respect to u.
对 u 进行积分。
Substitute back u = g(x) to express the result in terms of x.
代回 u = g(x)，把结果用 x 表示。
Add the constant of integration C for indefinite integrals.
对于不定积分，最后加上积分常数 C。

Always check your answer by differentiating it. The derivative should return the original integrand.

务必通过求导来检验答案，其导数应当回到原来的被积函数。

3. Type 1: Linear Substitution | 类型一：线性代换

The simplest type of substitution occurs when the integrand contains a linear expression such as ax + b. Let u = ax + b, so du = a dx, and solve for dx.

最简单的一类代换出现在被积函数含有线性表达式 ax + b 时。令 u = ax + b，则 du = a dx，再解出 dx 即可。

Consider the integral:

考虑以下积分：

∫ (2x + 1)³ dx

Let u = 2x + 1. Then du = 2 dx, so dx = du / 2. The integral becomes:

令 u = 2x + 1，则 du = 2 dx，因此 dx = du / 2。原积分变为：

∫ u³ · (du / 2) = (1/2) ∫ u³ du = u⁴ / 8 + C

Substituting back gives the final answer:

代回后得到最终结果：

(2x + 1)⁴ / 8 + C

4. Type 2: Integrands Containing a Function and Its Derivative | 类型二：被积函数含有函数及其导数

A very common pattern in IB exams is an integrand that contains both a function and its derivative, such as 2x and x², or cos x and sin x. This pattern signals a direct substitution.

IB考试中非常常见的一种模式是：被积函数同时包含某个函数及其导数，例如 2x 与 x²，或 cos x 与 sin x。这种模式提示我们应该直接进行代换。

For example, evaluate:

例如，计算：

∫ 2x cos(x²) dx

Here, the derivative of x² is 2x. Let u = x², so du = 2x dx. The integral becomes:

这里 x² 的导数为 2x。令 u = x²，则 du = 2x dx。积分变为：

∫ cos u du = sin u + C = sin(x²) + C

If the derivative appears only up to a constant factor, we can adjust by multiplying by an appropriate constant.

如果导数只相差一个常数倍数，我们可以通过乘以适当的常数来调整。

5. Type 3: Trigonometric Substitution | 类型三：三角换元

When an integrand contains expressions like √(a² – x²), √(a² + x²), or √(x² – a²), trigonometric substitution is often effective. This technique uses trigonometric identities to remove square roots.

当被积函数含有 √(a² – x²)、√(a² + x²) 或 √(x² – a²) 这类式子时，三角换元通常很有效。这一技巧利用三角恒等式去掉根号。

Example: evaluate:

例如，计算：

∫ √(1 – x²) dx

Let x = sin θ, so dx = cos θ dθ. Since √(1 – x²) = √(1 – sin² θ) = cos θ, the integral becomes:

令 x = sin θ，则 dx = cos θ dθ。因为 √(1 – x²) = √(1 – sin² θ) = cos θ，原积分变为：

∫ cos² θ dθ

Using the double-angle identity, cos² θ = (1 + cos 2θ) / 2, we integrate and convert back to x:

利用二倍角公式 cos² θ = (1 + cos 2θ) / 2，完成积分并代回 x：

(arcsin x) / 2 + x√(1 – x²) / 2 + C

6. Type 4: Substitution for Definite Integrals | 类型四：定积分的换元

For definite integrals, substitution can be performed without converting back to x. Instead, we change the limits of integration to match the new variable u.

对于定积分，换元时可以不代回 x，而是直接更新积分的上下限，使其与新变量 u 相匹配。

Example: evaluate:

例如，计算：

∫₀² x e^(x²) dx

Let u = x², so du = 2x dx, and x dx = du / 2. The old limits are x = 0 and x = 2, which become u = 0 and u = 4. Therefore:

令 u = x²，则 du = 2x dx，于是 x dx = du / 2。原积分的上下限 x = 0 和 x = 2 分别变为 u = 0 和 u = 4。因此：

(1/2) ∫₀⁴ e^u du = (e⁴ – 1) / 2

Notice that we did not need to express the final answer in terms of x again.

注意，这里我们不需要再把最终结果表示为 x 的函数。

7. Choosing the Right Substitution | 如何选择合适的代换

The most challenging part of substitution is choosing u. There are several useful strategies to guide this choice, especially under exam conditions.

换元法中最具挑战性的部分是如何选择 u。以下几个策略可以指导你作出选择，尤其是在考试条件下。

Look for an inner function whose derivative also appears in the integrand.
寻找一个“内部函数”，且其导数也出现在被积函数中。
If the integrand contains √(a² – x²), try x = a sin θ.
如果被积函数含有 √(a² – x²)，尝试 x = a sin θ。
If the integrand contains √(a² + x²), try x = a tan θ.
如果被积函数含有 √(a² + x²)，尝试 x = a tan θ。
If the integrand contains √(x² – a²), try x = a sec θ.
如果被积函数含有 √(x² – a²)，尝试 x = a sec θ。
For rational functions, try substituting the denominator or a part of the numerator.
对于有理函数，可尝试代换分母或分子的某一部分。

When unsure, try a substitution and check whether the resulting integral is simpler. If not, reconsider your choice.

如果不确定，不妨先尝试一种代换，并检查得到的积分是否更简单。如果不是，就重新考虑选择。

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

Students often lose marks due to small but avoidable errors. Knowing these pitfalls in advance can help you avoid them in the exam.

学生常常因为一些细小但可以避免的错误而失分。提前了解这些陷阱，可以帮助你在考试中避免它们。

Forgetting to replace dx correctly after substitution.
代换后忘记正确替换 dx。
Forgetting to change the limits of integration for definite integrals.
对于定积分，忘记更换积分上下限。
Omitting the constant of integration C in indefinite integrals.
在不定积分中漏写积分常数 C。
Choosing a substitution that makes the integral even more complicated.
选择了使积分更加复杂的代换。
Confusing u = g(x) with du = g'(x) dx; do not forget the derivative factor.
混淆 u = g(x) 与 du = g'(x) dx；不要忘记导数因子。

Always write down du explicitly and check whether every occurrence of x and dx has been replaced.

始终明确写出 du，并检查是否所有的 x 和 dx 都被替换完毕。

9. Practice Questions | 练习

Try these questions on your own before looking at the answers. They are representative of the types commonly seen in IB exams.

请先独立完成以下题目，再参考答案。这些题目代表了IB考试中常见的题型。

Question	Answer
Find ∫ 3x² (x³ + 1)⁴ dx	(x³ + 1)⁵ / 5 + C
Evaluate ∫₀¹ 2x / (1 + x²) dx	ln 2
Integrate ∫ sin³ x cos x dx	sin⁴ x / 4 + C

For the second question, let u = 1 + x², then du = 2x dx. The new limits are u = 1 and u = 2, giving ln 2 as the answer.

对于第二题，令 u = 1 + x²，则 du = 2x dx。新的上下限分别为 u = 1 和 u = 2，最终结果为 ln 2。

10. Conclusion and Exam Tips | 总结与考试建议

Integration by substitution is a versatile and essential method in IB Mathematics. Mastery of this technique requires practice, pattern recognition, and careful algebraic manipulation.

换元积分法是IB数学中一项通用而重要的方法。掌握这一技巧需要练习、模式识别以及细心的代数变形。

In the exam, always show your substitution steps clearly, write the differential du, and simplify systematically. For definite integrals, change the limits at the moment of substitution to avoid confusion.

在考试中，要清晰写出代换步骤，写清微分 du，并有条理地进行化简。对于定积分，应在代换的同时更新上下限，以避免混淆。

Finally, verify your indefinite integral by differentiation. This simple check can catch many errors and helps build confidence.

最后，通过求导来检验你的不定积分。这个简单的检查能发现许多错误，也能帮助你增强信心。

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IB Math: Applications and Techniques of Integration by Substitution | IB数学：换元积分法的应用与技巧