CE Board Exam Randomizer

⬅ Back to Integral Calculus Topics

Hyperbolic and Improper Integrals

Hyperbolic functions ($\sinh x$, $\cosh x$, $\tanh x$, …) behave like trig functions but are built from exponentials. They satisfy a Pythagorean-like identity $\cosh^2x - \sinh^2x = 1$. Their integrals mirror trig: $\int\sinh u\,du=\cosh u+C$, $\int\cosh u\,du=\sinh u+C$, $\int\operatorname{sech}^2 u\,du=\tanh u+C$, $\int\tanh u\,du=\ln(\cosh u)+C$. The most important engineering application is the catenary — the shape a flexible cable or chain takes under its own weight. Its equation is $y=a\cosh(x/a)$, where $a=H/w$ ($H$ = horizontal tension, $w$ = weight per unit length). The arc length from the low point over a half-span $b$ is $L=a\sinh(b/a)$, and the sag (vertical drop from the support) is $d=a\cosh(b/a)-a$.

$$\int\sinh u\,du=\cosh u+C,\quad\int\cosh u\,du=\sinh u+C,\quad\int\operatorname{sech}^2 u\,du=\tanh u+C$$
$$\text{Catenary: }y=a\cosh\!\left(\frac{x}{a}\right),\quad L=a\sinh\!\left(\frac{b}{a}\right),\quad d=a\!\left[\cosh\!\left(\frac{b}{a}\right)-1\right]$$

Hyperbolic Functions

$$\sinh x=\frac{e^x-e^{-x}}{2}$$
$$\cosh x=\frac{e^x+e^{-x}}{2}$$
$$\tanh x=\frac{\sinh x}{\cosh x}$$
$$\coth x=\frac{\cosh x}{\sinh x}$$
$$\operatorname{sech}x=\frac{1}{\cosh x}$$
$$\operatorname{csch}x=\frac{1}{\sinh x}$$

Key Hyperbolic Identities

$$\cosh^2 x-\sinh^2 x=1$$
$$1-\tanh^2 x=\operatorname{sech}^2 x$$
$$\coth^2 x-\operatorname{csch}^2 x=1$$

Derivatives of Hyperbolic Functions

$$\frac{d}{dx}\bigl(\sinh u\bigr)=\cosh u\,\frac{du}{dx}$$
$$\frac{d}{dx}\bigl(\cosh u\bigr)=\sinh u\,\frac{du}{dx}$$
$$\frac{d}{dx}\bigl(\tanh u\bigr)=\operatorname{sech}^2 u\,\frac{du}{dx}$$
$$\frac{d}{dx}\bigl(\coth u\bigr)=-\operatorname{csch}^2 u\,\frac{du}{dx}$$
$$\frac{d}{dx}\bigl(\operatorname{sech}u\bigr)=-\operatorname{sech}u\,\tanh u\,\frac{du}{dx}$$
$$\frac{d}{dx}\bigl(\operatorname{csch}u\bigr)=-\operatorname{csch}u\,\coth u\,\frac{du}{dx}$$

Improper Integrals

$$\int_a^\infty f(x)\,dx=\lim_{b\to\infty}\int_a^b f(x)\,dx$$
$$\int_{-\infty}^b f(x)\,dx=\lim_{a\to-\infty}\int_a^b f(x)\,dx$$
$$\int_a^b f(x)\,dx=\lim_{t\to c^-}\int_a^t f(x)\,dx+\lim_{t\to c^+}\int_t^b f(x)\,dx$$

Basic Hyperbolic Antiderivative

Evaluate $$\int \cosh(3x)\,dx.$$

Since $\frac{d}{dx}\sinh(3x)=3\cosh(3x)$,

$$\int \cosh(3x)\,dx=\boxed{\frac{1}{3}\sinh(3x)+C}$$

Exponential Improper Integral

Evaluate $$\int_0^\infty e^{-2x}\,dx.$$

Write the improper integral as a limit:

$$\int_0^\infty e^{-2x}\,dx=\lim_{b\to\infty}\left[-\frac{1}{2}e^{-2x}\right]_0^b$$

$$=0-\left(-\frac{1}{2}\right)=\boxed{\frac{1}{2}}$$

P-Integral That Converges

Evaluate $$\int_1^\infty \frac{dx}{x^3}.$$

Because $p=3>1$, the integral converges.

$$\int_1^\infty x^{-3}\,dx=\lim_{b\to\infty}\left[-\frac{1}{2x^2}\right]_1^b=\boxed{\frac{1}{2}}$$

Improper Integral With Vertical Asymptote

Evaluate $$\int_0^1 \frac{dx}{\sqrt{x}}.$$

The integrand is unbounded at $x=0$, so use a limit.

$$\int_0^1 x^{-1/2}\,dx=\lim_{a\to0^+}\left[2\sqrt{x}\right]_a^1=2$$

$$\boxed{2}$$

Harmonic Improper Integral Divergence

Determine whether $$\int_1^\infty \frac{dx}{x}$$ converges.

Compute the limit:

$$\int_1^\infty \frac{dx}{x}=\lim_{b\to\infty}\left[\ln x\right]_1^b=\lim_{b\to\infty}\ln b$$

The limit is infinite, so the integral diverges.

Integrating $\sinh^2 x$ Using an Identity

Evaluate $$\int\sinh^2 x\,dx.$$

Use the hyperbolic double-angle identity: $\cosh(2x)=\cosh^2x+\sinh^2x=1+2\sinh^2x$, so $\sinh^2x=\frac{\cosh2x-1}{2}$. This mirrors the trig power-reduction trick.

$$\int\sinh^2x\,dx=\int\frac{\cosh2x-1}{2}\,dx=\frac{\sinh2x}{4}-\frac{x}{2}+C$$

$$\boxed{\frac{\sinh2x}{4}-\frac{x}{2}+C}$$

Integrating $\tanh x$

Evaluate $$\int\tanh x\,dx.$$

Write $\tanh x=\frac{\sinh x}{\cosh x}$. The numerator is the derivative of the denominator.

$$\int\tanh x\,dx=\int\frac{\sinh x}{\cosh x}\,dx=\ln(\cosh x)+C$$

$$\boxed{\ln(\cosh x)+C}$$

Note: $\cosh x\ge1$ for all $x$, so no absolute value is needed.

Integration by Parts with a Hyperbolic Function

Evaluate $$\int x\sinh x\,dx.$$

Use IBP: let $u=x$, $dv=\sinh x\,dx$. Then $du=dx$, $v=\cosh x$.

$$\int x\sinh x\,dx=x\cosh x-\int\cosh x\,dx=x\cosh x-\sinh x+C$$

$$\boxed{x\cosh x-\sinh x+C}$$

Catenary Cable — Sag and Half-Length

A power line hangs between two towers 80 m apart (horizontal distance). The cable equation is $y=a\cosh(x/a)$ with $a=50$ m (low point at origin). Find: (a) the sag $d$ (how far the midpoint drops below the support level) and (b) the arc length of the cable on one side from the midpoint to a tower.

Half-span: $b=40$ m. Use $a=50$ m.

(a) Sag = height of supports minus height at midpoint. The midpoint is at $y=a$ (since $\cosh(0)=1$, $y_{\min}=a$). The supports are at $y=a\cosh(b/a)$.

$$d=a\cosh\!\left(\frac{b}{a}\right)-a=50\left[\cosh\!\left(\frac{40}{50}\right)-1\right]=50[\cosh(0.8)-1]$$

$$\cosh(0.8)=\frac{e^{0.8}+e^{-0.8}}{2}=\frac{2.2255+0.4493}{2}=1.3374$$

$$d=50(1.3374-1)=50(0.3374)=\boxed{16.87\text{ m}}$$

(b) Half-length:

$$L=a\sinh\!\left(\frac{b}{a}\right)=50\sinh(0.8)=50\cdot\frac{e^{0.8}-e^{-0.8}}{2}=50\cdot\frac{2.2255-0.4493}{2}=50(0.8881)=\boxed{44.40\text{ m}}$$

The total cable length between towers is $2\times44.40=88.80$ m, longer than the 80 m span.

Catenary Cable — Tension at Any Point

A flexible cable has a weight of $w=120$ N/m and hangs in a catenary with $a=H/w$. The horizontal tension is $H=6000$ N. (a) Find $a$. (b) Find the tension $T$ at a point where the sag places it at height $y=a\cosh(x/a)=a+5$ m above the lowest point.

(a) $a=H/w=6000/120=\boxed{50\text{ m}}$.

(b) The tension at any point equals $T=wy$ where $y$ is the height above the datum $y=0$ (lowest point set at $y=a$). Since the point is $5$ m above the lowest point, $y_{\text{actual}}=a+5=55$ m.

$$T=wy=120\times55=\boxed{6600\text{ N}}$$

Equivalently: $T=\sqrt{H^2+(ws)^2}$ where $s$ is the arc length, but the formula $T=wy$ (measuring $y$ from the catenary datum) gives the same result directly.

Catenary — Finding $a$ From Known Sag and Span

A transmission line has a horizontal span of 100 m and a known sag of 4 m. Find the catenary parameter $a$ and the total cable length.

Sag formula: $d=a\cosh(b/a)-a$ where $b=50$ m (half-span) and $d=4$ m.

$$4=a\left[\cosh\!\left(\frac{50}{a}\right)-1\right]$$

For small sag relative to span, the approximation $\cosh(t)\approx1+t^2/2$ is accurate:

$$4\approx a\cdot\frac{(50/a)^2}{2}=\frac{2500}{2a}=\frac{1250}{a}\quad\Rightarrow\quad a=\frac{1250}{4}=\boxed{312.5\text{ m}}$$

Total cable length:

$$S=2a\sinh\!\left(\frac{b}{a}\right)\approx2a\cdot\frac{b}{a}\left(1+\frac{b^2}{6a^2}\right)=2\cdot50\left(1+\frac{2500}{6\times312.5^2}\right)$$

$$\approx100\left(1+0.00427\right)=\boxed{100.43\text{ m}}$$

The cable is about 0.43 m longer than the span due to the sag.

Scroll to zoom