Calculus

Integration of Functions

Integration of Functions Logo

Integration of Hyperbolic Functions

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

Definitions of hyperbolic 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:

Derivatives and integrals of hyperbolic functions

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

Click or tap a problem to see the solution.

Example 1

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

Example 2

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

Example 3

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

Example 4

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

Example 5

Evaluate \[\int {{{\sinh }^3}xdx}.\]

Example 6

Evaluate the integral \[\int {x\sinh xdx}.\]

Example 7

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

Example 8

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

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.

Page 1 Page 2