FAC1004 L21-L22 — Integrals Involving Hyperbolic Functions

These lectures cover integration techniques for hyperbolic functions and integrals that lead to inverse hyperbolic functions. L21 focuses on basic hyperbolic integrals and $u$-substitution, while L22 covers standard forms producing inverse hyperbolic results and compares trigonometric versus hyperbolic substitution methods.

Learning Outcomes

  • L21: Understand how to integrate hyperbolic functions, apply standard integration formulas, and use basic identities and substitution techniques to solve foundational calculus problems.
  • L22: Understand how to integrate expressions that lead to inverse hyperbolic functions, using substitution and standard forms to obtain results in calculus problems.

L21 — Integrals of Hyperbolic Functions

Standard Integration Formulas (Theorem)

$$ \begin{aligned} \int \sinh u , du &= \cosh u + C \ \int \cosh u , du &= \sinh u + C \ \int \operatorname{sech}^2 u , du &= \tanh u + C \ \int \operatorname{csch}^2 u , du &= -\coth u + C \ \int \operatorname{sech} u \tanh u , du &= -\operatorname{sech} u + C \ \int \operatorname{csch} u \coth u , du &= -\operatorname{csch} u + C \end{aligned} $$

Worked Examples

Example 1 — Power rule with $u$-substitution

$$\int \sinh^5 x \cosh x , dx$$

Let $u = \sinh x$, then $du = \cosh x , dx$:

$$ \int u^5 , du = \frac{1}{6}u^6 + C = \frac{1}{6}\sinh^6 x + C $$

Example 2 — Integral of $\tanh x$

$$\int \tanh x , dx = \int \frac{\sinh x}{\cosh x}, dx$$

Let $u = \cosh x$, then $du = \sinh x , dx$:

$$ \int \frac{du}{u} = \ln u + C = \ln(\cosh x) + C $$

Example 3 — Radical with $u$-substitution

$$\int \sqrt{\tanh x} , \operatorname{sech}^2 x , dx$$

Let $u = \tanh x$, then $du = \operatorname{sech}^2 x , dx$:

$$ \int \sqrt{u}, du = \frac{u^{3/2}}{3/2} + C = \frac{2}{3}\tanh^{3/2} x + C $$

Example 4 — Negative coefficient from derivative

$$\int \coth^2 x , \operatorname{csch}^2 x , dx$$

Let $u = \coth x$, then $du = -\operatorname{csch}^2 x , dx$:

$$ \int -u^2 , du = -\frac{u^3}{3} + C = -\frac{1}{3}\coth^3 x + C $$

L22 — Integrals Involving Inverse Hyperbolic Functions

Motivation: Two Routes to the Same Result

The integral $\displaystyle\int \frac{dx}{\sqrt{a^2 + x^2}}$ can be evaluated by trigonometric substitution or hyperbolic substitution.

Route A — Trigonometric substitution

Let $x = a\tan\theta$, so $dx = a\sec^2\theta , d\theta$ and $a^2 + x^2 = a^2\sec^2\theta$.

$$ \int \frac{a\sec^2\theta , d\theta}{a\sec\theta} = \int \sec\theta , d\theta = \ln\bigl|\tan\theta + \sec\theta\bigr| + C $$

Substituting back:

$$ = \ln\left|\frac{x}{a} + \frac{\sqrt{a^2+x^2}}{a}\right| + C = \ln\bigl|x + \sqrt{a^2+x^2}\bigr| + D $$

where $D = C - \ln|a|$ absorbs the constant.

Route B — Hyperbolic substitution

Let $x = a\sinh u$, so $dx = a\cosh u , du$ and $a^2 + x^2 = a^2\cosh^2 u$.

$$ \int \frac{a\cosh u , du}{a\cosh u} = \int du = u + C = \sinh^{-1}!\left(\frac{x}{a}\right) + C $$

Both routes agree because:

$$ \int \frac{dx}{\sqrt{a^2 + x^2}} = \ln\bigl|x + \sqrt{a^2+x^2}\bigr| + D = \sinh^{-1}!\left(\frac{x}{a}\right) + C $$

Generalized Integration Theorem (Inverse Hyperbolic Forms)

$$ \begin{aligned} \int \frac{du}{\sqrt{a^2 + u^2}} &= \sinh^{-1}!\left(\frac{u}{a}\right) + C \ \int \frac{du}{\sqrt{u^2 - a^2}} &= \cosh^{-1}!\left(\frac{u}{a}\right) + C \ \int \frac{du}{a^2 - u^2} &= \begin{cases} \displaystyle\frac{1}{a}\tanh^{-1}!\left(\frac{u}{a}\right) + C, & |u| < a \ \displaystyle\frac{1}{a}\coth^{-1}!\left(\frac{u}{a}\right) + C, & |u| > a \end{cases} \ \int \frac{du}{u\sqrt{a^2 + u^2}} &= -\frac{1}{a}\operatorname{csch}^{-1}!\left(\frac{u}{a}\right) + C \ \int \frac{du}{u\sqrt{a^2 - u^2}} &= -\frac{1}{a}\operatorname{sech}^{-1}!\left(\frac{u}{a}\right) + C \end{aligned} $$

Worked Examples

Example 1 — Form $\displaystyle\int \frac{dx}{\sqrt{a^2+u^2}}$

$$\int \frac{dx}{\sqrt{1 + 9x^2}}$$

Let $u = 3x$, so $du = 3,dx$ and $dx = \frac{du}{3}$. Then $a = 1$:

$$ \int \frac{du}{3\sqrt{1^2 + u^2}} = \frac{1}{3}\sinh^{-1}!\left(\frac{u}{1}\right) + C = \frac{1}{3}\sinh^{-1}(3x) + C $$

Example 2 — Form $\displaystyle\int \frac{dx}{x\sqrt{a^2+u^2}}$

$$\int \frac{dx}{x\sqrt{9 + 4x^2}}$$

Let $u = 2x$, so $x = \frac{u}{2}$ and $dx = \frac{du}{2}$. Then $a = 3$:

$$ \int \frac{\frac{du}{2}}{\frac{u}{2}\sqrt{3^2 + u^2}} = \int \frac{du}{u\sqrt{3^2 + u^2}} = -\frac{1}{3}\operatorname{csch}^{-1}!\left(\frac{u}{3}\right) + C = -\frac{1}{3}\operatorname{csch}^{-1}!\left(\frac{2x}{3}\right) + C $$

Summary

Technique When to Use
Direct hyperbolic integral Integrand matches standard form exactly
$u$-substitution One hyperbolic factor is the derivative of another
Hyperbolic substitution Radical forms $\sqrt{a^2+x^2}$ or $\sqrt{x^2-a^2}$
Standard inverse forms After $u$-substitution reduces to $\frac{1}{\sqrt{a^2\pm u^2}}$, $\frac{1}{a^2-u^2}$, etc.

The piecewise result for $\displaystyle\int \frac{du}{a^2-u^2}$ is essential: use $\tanh^{-1}$ when $|u|<a$ and $\coth^{-1}$ when $|u|>a$.

Related

Source File

LECTURE_NOTES_2526/L21 L22 Inverse Hyperbolic function 2 - full version.pdf