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