Calculus

Integration of Functions

Integration of Functions Logo

Integration of Hyperbolic Functions

  • 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

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

    Page 1
    Problems 1-2
    Page 2
    Problems 3-8