Integration of Hyperbolic Functions | 双曲函数的积分

📚 Integration of Hyperbolic Functions | 双曲函数的积分

Hyperbolic functions appear frequently in A-Level Further Mathematics, and integrating them requires a confident grasp of their definitions, identities, and the reciprocal relationships they share with exponential and trigonometric functions.

双曲函数在 A-Level 进阶数学中频繁出现，积分双曲函数需要扎实掌握其定义、恒等式以及与指数函数和三角函数之间的对偶关系。

1. Definitions and Derivatives You Must Know | 必须熟记的定义与导数

Recall the exponential definitions, since nearly every integration technique for hyperbolics ultimately rests on them: sinh x = (eˣ − e⁻ˣ)/2 and cosh x = (eˣ + e⁻ˣ)/2. From these, the derivatives follow immediately:

回顾指数定义至关重要，因为几乎所有双曲积分技巧最终都建立在这些定义之上：sinh x = (eˣ − e⁻ˣ)/2，cosh x = (eˣ + e⁻ˣ)/2。由此可立即得到导数关系：

d/dx (sinh x) = cosh x, d/dx (cosh x) = sinh x, d/dx (tanh x) = sech²x

Because the derivative of cosh x is positive sinh x (not negative, as with cos x), integration of hyperbolic functions is often simpler than the trigonometric counterpart — no sign changes are introduced.

由于 cosh x 的导数为正 sinh x（与 cos x 的负号不同），双曲函数的积分通常比三角函数的积分更简单——不会引入符号变化。

2. Basic Integrals: sinh, cosh, sech² | 基本积分：sinh、cosh、sech²

The three core integrals follow directly from the derivatives above. These must be memorised and applied without hesitation:

三个核心积分直接来自上述导数，必须牢记并能毫不犹豫地运用：

∫ sinh x dx = cosh x + C
∫ cosh x dx = sinh x + C
∫ sech²x dx = tanh x + C

Notice that no negative signs appear, unlike ∫ sin x dx = −cos x + C. This symmetry makes hyperbolic integration less error-prone once the basic forms are internalised.

注意此处不出现负号，不像 ∫ sin x dx = −cos x + C。一旦掌握了基本形式，这种对称性使得双曲积分更不容易出错。

3. Integrating tanh x via Substitution | 利用代换积分 tanh x

The integral of tanh x is not a standard derivative, so we rewrite it as a ratio and use the substitution u = cosh x:

tanh x 的积分不是标准导数，因此我们将其改写为比值形式，并令 u = cosh x 进行代换：

∫ tanh x dx = ∫ (sinh x / cosh x) dx = ln|cosh x| + C

Since cosh x ≥ 1 for all real x, the absolute value is often omitted, giving ∫ tanh x dx = ln(cosh x) + C. The same method works for ∫ coth x dx = ln|sinh x| + C.

由于对所有实数 x 有 cosh x ≥ 1，绝对值常可省略，得到 ∫ tanh x dx = ln(cosh x) + C。同样的方法可求 ∫ coth x dx = ln|sinh x| + C。

4. Key Hyperbolic Identities for Integration | 积分类键恒等式

Just as with trigonometric integrals, double-angle and Pythagorean-style identities allow us to rewrite products and powers into integrable linear combinations:

与三角积分一样，二倍角和毕达哥拉斯型恒等式允许我们将乘积和幂次改写为可积的线性组合：

cosh²x − sinh²x = 1 | 基本平方关系
cosh 2x = cosh²x + sinh²x = 2cosh²x − 1 = 2sinh²x + 1 | 二倍角公式
sinh 2x = 2sinh x cosh x | 二倍角公式

From cosh 2x = 2cosh²x − 1, we obtain cosh²x = (cosh 2x + 1)/2. This mirrors cos²x = (cos 2x + 1)/2, enabling direct integration of even powers.

由 cosh 2x = 2cosh²x − 1 可得 cosh²x = (cosh 2x + 1)/2。这与 cos²x = (cos 2x + 1)/2 相对应，从而可以直接积分偶次幂。

5. Integrating Powers: sinh²x and Related Forms | 幂次积分：sinh²x 及相关形式

To integrate sinh²x, use the identity sinh²x = (cosh 2x − 1)/2:

要积分 sinh²x，使用恒等式 sinh²x = (cosh 2x − 1)/2：

∫ sinh²x dx = ∫ (cosh 2x − 1)/2 dx = (1/2)[(sinh 2x)/2 − x] + C

Similarly, ∫ cosh²x dx = (1/2)[(sinh 2x)/2 + x] + C. For odd powers such as sinh³x, factor one sinh x and use cosh²x − sinh²x = 1 to convert the remainder, then substitute u = cosh x.

类似地，∫ cosh²x dx = (1/2)[(sinh 2x)/2 + x] + C。对于奇次幂如 sinh³x，提取一个 sinh x，用 cosh²x − sinh²x = 1 转换剩余部分，然后令 u = cosh x 代换。

6. Integration by Substitution with Hyperbolic Functions | 双曲函数代换积分法

When the integrand contains an expression of the form √(x² + a²), √(x² − a²), or √(a² − x²), hyperbolic substitution is often more elegant than trigonometric substitution:

当被积函数含有 √(x² + a²)、√(x² − a²) 或 √(a² − x²) 形式时，双曲代换通常比三角代换更为优雅：

For √(x² + a²): set x = a sinh t, using cosh²t − sinh²t = 1
For √(x² − a²): set x = a cosh t, using cosh²t − sinh²t = 1
For √(a² − x²): set x = a tanh t, using 1 − tanh²t = sech²t

For example, ∫ dx/√(x² + 1) with x = sinh t gives dx = cosh t dt and √(x² + 1) = cosh t, so the integral simplifies to ∫ dt = t = arsinh x + C.

例如，∫ dx/√(x² + 1)，令 x = sinh t，则 dx = cosh t dt，√(x² + 1) = cosh t，积分简化为 ∫ dt = t = arsinh x + C。

7. Inverse Hyperbolic Functions as Results | 积分结果中的反双曲函数

The derivatives of inverse hyperbolic functions produce three standard integrals that appear throughout AQA exam papers:

反双曲函数的导数产生了三个贯穿 AQA 考卷的标准积分：

∫ dx/√(x² + a²) = arsinh(x/a) + C

∫ dx/√(x² − a²) = arcosh(x/a) + C

∫ dx/(a² − x²) = (1/a) artanh(x/a) + C

These are often presented in logarithmic form as well: arsinh x = ln(x + √(x² + 1)) and artanh x = ½ ln((1 + x)/(1 − x)). Both forms are acceptable in examinations — check the question’s required form.

这些也常以对数形式给出：arsinh x = ln(x + √(x² + 1))，artanh x = ½ ln((1 + x)/(1 − x))。两种形式在考试中均可接受——请注意题目的形式要求。

8. Completing the Square for Quadratic Denominators | 二次分母的配方法

Whenever the integrand contains a quadratic denominator or a square root of a quadratic, complete the square first to match the standard inverse hyperbolic forms:

当被积函数含有二次分母或二次根式时，应首先配方以匹配标准反双曲形式：

∫ dx/√(x² + 4x + 5) = ∫ dx/√((x + 2)² + 1) = arsinh(x + 2) + C

Here x² + 4x + 5 = (x + 2)² + 1, so the integral matches the form ∫ du/√(u² + 1) with u = x + 2. This technique combines algebraic manipulation with the standard results above.

此处 x² + 4x + 5 = (x + 2)² + 1，积分符合 ∫ du/√(u² + 1) 的形式，其中 u = x + 2。该技巧将代数操作与上述标准结果相结合。

9. Definite Integrals with Hyperbolic Functions | 双曲函数的定积分

For definite integrals, always evaluate the antiderivative at the limits using exact values of sinh and cosh at common arguments. For instance:

对于定积分，务必在积分限处用 sinh 和 cosh 的精确值计算原函数。例如：

∫₀¹ cosh x dx = [sinh x]₀¹ = sinh 1 = (e − e⁻¹)/2

When inverse hyperbolic functions appear in the antiderivative, substitute the limits carefully and simplify using logarithmic laws, especially when the limits are 0 and a positive constant.

当原函数中出现反双曲函数时，应仔细代入上下限并利用对数法则化简，特别是当上下限为 0 和某个正常数时。

10. Exam-Style Strategies and Common Pitfalls | 考试策略与常见误区

AQA exam questions often combine hyperbolic integration with other topics, such as solving differential equations or finding areas. A reliable strategy is to identify whether the integrand resolves to a derivative of a hyperbolic function, requires an identity to simplify, or matches a standard inverse hyperbolic form.

AQA 考题常将双曲积分与其他主题结合，如求解微分方程或求面积。可靠的策略是判断被积函数是某个双曲函数的导数、需要恒等式简化、还是匹配某种标准反双曲形式。

Pitfall 1: Forgetting the factor 1/a in ∫ dx/(a² − x²) = (1/a) artanh(x/a) + C.
Pitfall 2: Confusing ∫ sech²x dx with ∫ sech x dx; the latter requires a substitution or identity and is not a standard derivative.
Pitfall 3: Omitting the constant of integration in indefinite integrals.
Pitfall 4: Using the circular trigonometric versions of identities rather than the hyperbolic forms, especially sign differences in double-angle formulas.

误区一：忘记 ∫ dx/(a² − x²) = (1/a) artanh(x/a) + C 中的因子 1/a。

误区二：混淆 ∫ sech²x dx 与 ∫ sech x dx；后者需要代换或恒等式，不是标准导数。

误区三：不定积分中遗漏积分常数 C。

误区四：误用三角恒等式版本而非双曲版本，尤其是二倍角公式中的符号差异。

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Integration of Hyperbolic Functions | 双曲函数的积分