# Integration of Hyperbolic Functions

The hyperbolic functions are defined in terms of the exponential functions:

The hyperbolic functions have identities that are similar to those of trigonometric functions:

${\cosh ^2}x - {\sinh ^2}x = 1;$
$1 - {\tanh ^2}x = {\text{sech}^2}x;$
${\coth ^2}x - 1 = {\text{csch}^2}x;$
$\sinh 2x = 2\sinh x\cosh x;$
$\cosh 2x = {\cosh ^2}x + {\sinh ^2}x.$

Since the hyperbolic functions are expressed in terms of $${e^x}$$ and $${e^{ - x}},$$ we can easily derive rules for their differentiation and integration:

In certain cases, the integrals of hyperbolic functions can be evaluated using the substitution

$u = {e^x},\;\; \Rightarrow x = \ln u,\;\; dx = \frac{{du}}{u}.$

## Solved Problems

### Example 1.

Calculate the integral $\int {{\frac{{\cosh x}}{{2 + 3\sinh x}}} dx}.$

Solution.

We make the substitution: $$u = 2 + 3\sinh x,$$ $$du = 3\cosh x dx.$$ Then $$\cosh x dx = {\frac{{du}}{3}}.$$ Hence, the integral is

$\int {\frac{{\cosh x}}{{2 + 3\sinh x}}dx} = \int {\frac{{\frac{{du}}{3}}}{u}} = \frac{1}{3}\int {\frac{{du}}{u}} = \frac{1}{3}\ln \left| u \right| + C = \frac{1}{3}\ln \left| {2 + 3\sinh x} \right| + C.$

### Example 2.

Evaluate the integral $\int {\frac{{\sinh x}}{{1 + \cosh x}} dx}.$

Solution.

Using the substitution

$u = 1 + \cosh x,\;\; du = \sinh xdx,$

we get

$I = \int {\frac{{\sinh x}}{{1 + \cosh x}}dx} = \int {\frac{{du}}{u}} = \ln \left| u \right| + C = \ln \left| {1 + \cosh x} \right| + C.$

The hyperbolic cosine is a positive function. Hence, we can write the answer in the form

$I = \ln \left( {1 + \cosh x} \right) + C.$

### Example 3.

Evaluate the integral $\int {{{\sinh }^2}xdx}.$

Solution.

Combining the identities

${\cosh ^2}x - {\sinh ^2}x = 1,$
$\cosh 2x = {\cosh ^2}x + {\sinh ^2}x,$

we write:

$\cosh 2x = {\cosh ^2}x + {\sinh ^2}x = 1 + {\sinh ^2}x + {\sinh ^2}x = 1 + 2\,{\sinh ^2}x.$

So we can use the following half-angle formula:

${\sinh ^2}x = \frac{1}{2}\left( {\cosh 2x - 1} \right).$

Then the integral becomes

$\int {{{\sinh }^2}xdx} = \frac{1}{2}\int {\left( {\cosh 2x - 1} \right)dx} = \frac{1}{2}\left( {\frac{{\sinh 2x}}{2} - x} \right) + C = \frac{1}{4}\sinh 2x - \frac{x}{2} + C.$

### Example 4.

Evaluate the integral $\int {{{\cosh }^2}xdx}.$

Solution.

We reduce the power of the integrand using the identities

${\cosh ^2}x - {\sinh ^2}x = 1,$
$\cosh 2x = {\cosh ^2}x + {\sinh ^2}x.$

Then

$\cosh 2x = {\cosh ^2}x + {\sinh ^2}x = {\cosh ^2}x + {\cosh ^2}x - 1 = 2{\cosh ^2}x - 1,$

and

${\cosh ^2}x = \frac{1}{2}\left( {\cosh 2x + 1} \right).$

Now we can find the initial integral:

$\int {{{\cosh }^2}xdx} = \frac{1}{2}\int {\left( {\cosh 2x + 1} \right)dx} = \frac{1}{2}\left( {\frac{{\sinh 2x}}{2} + x} \right) + C = \frac{1}{4}\sinh 2x + \frac{x}{2} + C.$

### Example 5.

Evaluate $\int {{{\sinh }^3}xdx}.$

Solution.

Since

${\cosh ^2}x - {\sinh ^2}x = 1,$

and hence,

${\sinh^2}x = {\cosh ^2}x - 1,$

we can write the integral as

$I = \int {{{\sinh }^3}xdx} = \int {{{\sinh }^2}x\sinh xdx} = \int {\left( {{\cosh^2}x - 1} \right)\sinh xdx} .$

Making the substitution $$u = \cosh x,$$ $$du = \sinh xdx,$$ we obtain

$I = \int {\left( {{\cosh^2}x - 1} \right)\sinh xdx} = \int {\left( {{u^2} - 1} \right)du} = \frac{{{u^3}}}{3} - u + C = \frac{{{{\cosh }^3}x}}{3} - \cosh x + C.$

### Example 6.

Evaluate the integral $\int {x\sinh xdx}.$

Solution.

We use integration by parts:

$\int {udv} = uv - \int{vdu} .$

Let $$u = x,$$ $$dv=\sinh xdx.$$ Then

$du = dx,\;v = \int {\sinh xdx} = \cosh x.$

Hence, the integral is

$\int {x\sinh xdx} = x\cosh x - \int {\cosh xdx} = x\cosh x - \sinh x + C.$

### Example 7.

Evaluate the integral $\int {{e^x}\sinh xdx}.$

Solution.

Since

$\sinh x = {\frac{{{e^x} - {e^{ - x}}}}{2}},$

we obtain

$\int {{e^x}\sinh xdx} = \int {{e^x}\frac{{{e^x} - {e^{ - x}}}}{2}dx} = \frac{1}{2}\int {\left( {{e^{2x}} - 1} \right)dx} = \frac{1}{2}\left( {\frac{1}{2}{e^{2x}} - x} \right) + C = \frac{{{e^{2x}}}}{4} - \frac{x}{2} + C.$

### Example 8.

Evaluate the integral $\int {{e^{2x}}\cosh xdx}.$

Solution.

By definition,

$\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}.$

Hence,

$\int {{e^{2x}}\cosh xdx} = \int {{e^{2x}}\left( {\frac{{{e^x} + {e^{ - x}}}}{2}} \right)dx} = \frac{1}{2}\int {\left( {{e^{3x}} + {e^x}} \right)dx} = \frac{1}{2}\left( {\frac{{{e^{3x}}}}{3} + {e^x}} \right) + C = \frac{{{e^{3x}}}}{6} + \frac{{{e^x}}}{2} + C.$

See more problems on Page 2.