The \(6\) basic hyperbolic functions are defined by
| \(\sinh x = \large\frac{{{e^x} – {e^{ – x}}}}{2}\normalsize\) | \(\cosh x = \large\frac{{{e^x} + {e^{ – x}}}}{2}\normalsize\) |
| \(\tanh x = \large\frac{{\sinh x}}{{\cosh x}}\normalsize = \large\frac{{{e^x} – {e^{ – x}}}}{{{e^x} + {e^{ – x}}}}\normalsize\) | \(\coth x = \large\frac{{\cosh x}}{{\sinh x}}\normalsize = \large\frac{{{e^x} + {e^{ – x}}}}{{{e^x} – {e^{ – x}}}}\normalsize\) |
| \(\text{sech}\,x = \large\frac{1}{{\cosh x}}\normalsize = \large\frac{2}{{{e^x} + {e^{ – x}}}}\normalsize\) | \(\text{csch}\,x = \large\frac{1}{{\sinh x}}\normalsize = \large\frac{2}{{{e^x} – {e^{ – x}}}}\normalsize\) |
| \(\sinh x = \large\frac{{{e^x} – {e^{ – x}}}}{2}\normalsize\) |
| \(\cosh x = \large\frac{{{e^x} + {e^{ – x}}}}{2}\normalsize\) |
| \(\tanh x = \large\frac{{\sinh x}}{{\cosh x}}\normalsize = \large\frac{{{e^x} – {e^{ – x}}}}{{{e^x} + {e^{ – x}}}}\normalsize\) |
| \(\coth x = \large\frac{{\cosh x}}{{\sinh x}}\normalsize = \large\frac{{{e^x} + {e^{ – x}}}}{{{e^x} – {e^{ – x}}}}\normalsize\) |
| \(\text{sech}\,x = \large\frac{1}{{\cosh x}}\normalsize = \large\frac{2}{{{e^x} + {e^{ – x}}}}\normalsize\) |
| \(\text{csch}\,x = \large\frac{1}{{\sinh x}}\normalsize = \large\frac{2}{{{e^x} – {e^{ – x}}}}\normalsize\) |
There are the following differentiation and integration formulas for hyperbolic functions:
| \({\left( {\sinh x} \right)^\prime } = \cosh x\) | \({\large\int\normalsize} {\cosh x dx} = \sinh x + C\) |
| \({\left( {\cosh x} \right)^\prime } = \sinh x\) | \({\large\int\normalsize} {\sinh x dx} = \cosh x + C\) |
| \({\left( {\tanh x} \right)^\prime } = {\text{sech}^2}x\) | \({\large\int\normalsize} {{\text{sech}^2}x dx} = \tanh x + C\) |
| \({\left( {\coth x} \right)^\prime } = -{\text{csch}^2}x\) | \({\large\int\normalsize} {{\text{csch}^2}x dx} = -\coth x + C\) |
| \({\left( {\text{sech}\,x} \right)^\prime } \) \(= – \text{sech}\,x\tanh x\) | \({\large\int\normalsize} {\text{sech}\,x\tanh xdx} \) \( = – \text{sech}\,x + C\) |
| \({\left( {\text{csch}\,x} \right)^\prime } \) \(= – \text{csch}\,x\coth x\) | \({\large\int\normalsize} {\text{csch}\,x\coth xdx} \) \(= – \text{csch}\,x + C\) |
| \({\left( {\sinh x} \right)^\prime } = \cosh x\) |
| \({\large\int\normalsize} {\cosh x dx} = \sinh x + C\) |
| \({\left( {\cosh x} \right)^\prime } = \sinh x\) |
| \({\large\int\normalsize} {\sinh x dx} = \cosh x + C\) |
| \({\left( {\tanh x} \right)^\prime } = {\text{sech}^2}x\) |
| \({\large\int\normalsize} {{\text{sech}^2}x dx} = \tanh x + C\) |
| \({\left( {\coth x} \right)^\prime } = -{\text{csch}^2}x\) |
| \({\large\int\normalsize} {{\text{csch}^2}x dx} = -\coth x + C\) |
| \({\left( {\text{sech}\,x} \right)^\prime } \) \(= – \text{sech}\,x\tanh x\) |
| \({\large\int\normalsize} {\text{sech}\,x\tanh xdx} \) \( = – \text{sech}\,x + C\) |
| \({\left( {\text{csch}\,x} \right)^\prime } \) \(= – \text{csch}\,x\coth x\) |
| \({\large\int\normalsize} {\text{csch}\,x\coth xdx} \) \( = – \text{csch}\,x + C\) |
We provide here a list of useful hyperbolic identities:
- \({\cosh ^2}x – {\sinh ^2}x\) \(= 1\)
- \(\sinh 2x \) \(= 2\sinh x\cosh x\)
- \(\cosh 2x \) \(= {\cosh ^2}x + {\sinh ^2}x\)
When an integrand contains a hyperbolic function, the integral can be reduced to integrating a rational function by using the substitution \(u = {e^x},\) \(x = \ln u,\) \(dx = {\large\frac{{du}}{u}\normalsize}.\)
Solved Problems
Click a problem to see the solution.
Example 1
Calculate the integral \({\large\int\normalsize} {{\large\frac{{\cosh x}}{{2 + 3\sinh x}}\normalsize} dx}.\)Example 2
Evaluate \({\large\int\normalsize} {{{\sinh }^3}xdx}.\)Example 3
Evaluate the integral \({\large\int\normalsize} {x\sinh xdx}.\)Example 4
Evaluate the integral \({\large\int\normalsize} {{e^x}\sinh xdx}.\)Example 5
Find the integral \({\large\int\normalsize} {\large\frac{{dx}}{{1 + \cosh x}}\normalsize}.\)Example 6
Find the integral \({\large\int\normalsize} {\large\frac{{dx}}{{\sinh x + 2\cosh x}}\normalsize}.\)Example 7
Calculate the integral \({\large\int\normalsize} {\sinh 2x\cosh 3xdx}.\)Example 8
Evaluate the integral \({\large\int\normalsize} {\sinh x\cos xdx}.\)Example 1.
Calculate the integral \({\large\int\normalsize} {{\large\frac{{\cosh x}}{{2 + 3\sinh x}}\normalsize} dx}.\)Solution.
We make the substitution: \(u = 2 + 3\sinh x,\) \(du = 3\cosh x dx.\) Then \(\cosh x dx = {\large\frac{{du}}{3}\normalsize}.\) 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 \({\large\int\normalsize} {{{\sinh }^3}xdx}.\)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
\[
{I = \int {\left( {{\cosh^2}x – 1} \right)\sinh xdx} }
= {\int {\left( {{u^2} – 1} \right)du} }
= {\frac{{{u^3}}}{3} – u + C }
= {\frac{{{{\cosh }^3}x}}{3} – \cosh x }+{ C.}
\]