Integrals of Hyperbolic Functions

Hyperbolic functions, similar to trigonometric functions, are expressed in terms of exponential functions (e^x and e^-x). However, they are based on hyperbolas rather than circles. To compute their integrals, we apply standard integration techniques.

Before we understand the integration of the hyperbolic functions, here is a summary of their derivatives.

• ${\dfrac{d}{dx}\left( \sinh x\right) =\cosh x}$
• ${\dfrac{d}{dx}\left( \cosh x\right) =\sinh x}$
• ${\dfrac{d}{dx}\left( \tanh x\right) =sech ^{2}x}$
• ${\dfrac{d}{dx}\left( cosech \  x\right) =-cosech \  x\cdot \coth x}$
• ${\dfrac{d}{dx}\left( sech \  x\right) =-sech \  x\cdot \tanh x}$
• ${\dfrac{d}{dx}\left( \coth x\right) =-cosech ^{2}x}$

Since integration (antidifferentiation) is the reverse of differentiation, we can better understand how integrals return us to familiar functions by looking at the corresponding derivatives.

• ${\int \cosh x \  dx=\sinh x+C}$
• ${\int \sinh x \  dx=\cosh x+C}$
• ${\int sech^{2}x \  dx=\tanh x+C}$
• ${\int cosech \  x\coth x \  dx=-cosech \  x+C}$
• ${\int sech \  x\tanh x \  dx=-sech \  x+C}$
• ${\int cosech^{2}x \  dx=-\coth x+C}$, here, C is the integrating constant. 

The integrals of these hyperbolic functions can be evaluated using the substitution:

e^x = u ⇒ x = lnu ⇒ ${dx=\dfrac{du}{u}}$, which simplifies the integration process.

Integrating the basic hyperbolic functions plays an important role in a wide range of applications, such as in the solution of differential equations and the modeling of physical systems.

Formulas

The integrals of the basic hyperbolic functions are listed: 

• sinh x: ${\int \sinh x \  dx=\cosh x+C}$
• cosh x: ${\int \cosh x \  dx=\sinh x+C}$
• tanh x: ${\int \tanh x \  dx=\ln \left| \cosh x\right| +C}$
• cosech x: ${\int cosech \  x \  dx=\ln \left( \tanh \left( \dfrac{x}{2}\right) \right) +C}$
• sech x: ${\int sech \  x \  dx=\tan ^{-1}\left( \sinh x\right) +C}$
• coth x: ${\int \coth x \  dx=\ln \left| \sinh x\right| +C}$

To calculate these integrals, we use standard integration methods.

Solved Examples

Evaluate the integral of sinh² x

Solution:

${\int \cosh ^{2}xdx}$
= ${\int \dfrac{1-\cosh 2x}{2}dx}$
= ${\dfrac{1}{2}\left[ \int 1dx-\int \cosh 2xdx\right]}$
= ${\dfrac{1}{2}\left[ x-\dfrac{\sinh 2x}{2}\right] +C}$
= ${\dfrac{x}{2}-\dfrac{\sinh 2x}{4}+C}$, here, C = integrating constant

Compute the integral ${\int \dfrac{\sinh x}{3+2\cosh x}dx}$

Solution:

${\int \dfrac{\sinh x}{3+2\cosh x}dx}$ …..(i)
Using the substitution u = sinh x, we get ${\dfrac{du}{dx}=\cosh x}$
⇒ du = cosh x dx
Also, ${\cosh x=\sqrt{1+\sinh ^{2}x}}$ ⇒ ${\cosh x=\sqrt{1+u^{2}}}$
Now, from (i)
${\int \dfrac{\sinh x}{3+2\cosh x}dx}$
= ${\int \dfrac{u}{3+2\sqrt{1+u^{2}}}\cdot \dfrac{du}{\sqrt{1+u^{2}}}}$
= ${\int \dfrac{udu}{3\sqrt{1+u^{2}}+2\left( 1+u^{2}\right) }}$ …..(ii)
Let ${\sqrt{1+u^{2}}=t}$ ⇒ 1 + u² = t² ⇒ u du = t dt
Now, from (ii)
= ${\int \dfrac{tdt}{3t+2t^{2}}}$
= ${\int \dfrac{tdt}{t\left( 3+2t\right) }}$
= ${\int \dfrac{dt}{\left( 3+2t\right) }}$
= ${\dfrac{1}{2}\ln \left| 3+2t\right| +C}$
= ${\dfrac{1}{2}\ln \left| 3+2\sqrt{1+u^{2}}\right| +C}$
= ${\dfrac{1}{2}\ln \left| 3+2\cosh x\right| +C}$, here, C = integrating constant

Evaluate: ${\int \dfrac{\tanh x}{1-\tanh ^{2}x}dx}$

Solution:

${\int \dfrac{\tanh x}{1-\tanh ^{2}x}dx}$
= ${\int \dfrac{\tanh x}{sech^{2}x}dx}$
= ${\int \dfrac{\dfrac{\sinh x}{\cosh x}}{\dfrac{1}{\cosh ^{2}x}}dx}$
= ${\int \dfrac{\sinh x}{\cosh x}\cdot \cosh ^{2}xdx}$
= ${\int \sinh x\cdot \cosh xdx}$
= ${\dfrac{1}{2}\int 2\sinh x\cdot \cosh xdx}$
= ${\dfrac{1}{2}\int \sinh 2xdx}$
= ${\dfrac{1}{2}\left[ \dfrac{\cosh 2x}{2}\right] +C}$
= ${\dfrac{\cosh 2x}{4}+C}$, here, C = integrating constant

Find the integral of ${\int x\cosh xdx}$

Solution:

As we know, from the integration by parts, ${\int udv=uv-\int vdu}$
Let u = x and dv = cosh x dx, then, du = dx and v = ${\int \cosh xdx}$ = sinh x
Given, ${\int x\cosh xdx}$
= x sinh x – ${\int \sinh xdx}$
= x sinh x – cosh x + C, here, C = integrating constant