Calculus

Integration of Functions

Integration of Functions Logo

Integration of Hyperbolic Functions

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

    \(\sinh x = \large\frac{{{e^x} – {e^{ – x}}}}{2}\normalsize\)\(\cosh x = \large\frac{{{e^x} + {e^{ – x}}}}{2}\normalsize\)
    \(\tanh x = \large\frac{{\sinh x}}{{\cosh x}}\normalsize = \large\frac{{{e^x} – {e^{ – x}}}}{{{e^x} + {e^{ – x}}}}\normalsize\)\(\coth x = \large\frac{{\cosh x}}{{\sinh x}}\normalsize = \large\frac{{{e^x} + {e^{ – x}}}}{{{e^x} – {e^{ – x}}}}\normalsize\)
    \(\text{sech}\,x = \large\frac{1}{{\cosh x}}\normalsize = \large\frac{2}{{{e^x} + {e^{ – x}}}}\normalsize\)\(\text{csch}\,x = \large\frac{1}{{\sinh x}}\normalsize = \large\frac{2}{{{e^x} – {e^{ – x}}}}\normalsize\)
    \(\sinh x = \large\frac{{{e^x} – {e^{ – x}}}}{2}\normalsize\)
    \(\cosh x = \large\frac{{{e^x} + {e^{ – x}}}}{2}\normalsize\)
    \(\tanh x = \large\frac{{\sinh x}}{{\cosh x}}\normalsize = \large\frac{{{e^x} – {e^{ – x}}}}{{{e^x} + {e^{ – x}}}}\normalsize\)
    \(\coth x = \large\frac{{\cosh x}}{{\sinh x}}\normalsize = \large\frac{{{e^x} + {e^{ – x}}}}{{{e^x} – {e^{ – x}}}}\normalsize\)
    \(\text{sech}\,x = \large\frac{1}{{\cosh x}}\normalsize = \large\frac{2}{{{e^x} + {e^{ – x}}}}\normalsize\)
    \(\text{csch}\,x = \large\frac{1}{{\sinh x}}\normalsize = \large\frac{2}{{{e^x} – {e^{ – x}}}}\normalsize\)

    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:

    \({\left( {\sinh x} \right)^\prime } = \cosh x\)\({\large\int\normalsize} {\cosh x dx} = \sinh x + C\)
    \({\left( {\cosh x} \right)^\prime } = \sinh x\)\({\large\int\normalsize} {\sinh x dx} = \cosh x + C\)
    \({\left( {\tanh x} \right)^\prime } = {\text{sech}^2}x\)\({\large\int\normalsize} {{\text{sech}^2}x dx} = \tanh x + C\)
    \({\left( {\coth x} \right)^\prime } = -{\text{csch}^2}x\)\({\large\int\normalsize} {{\text{csch}^2}x dx} = -\coth x + C\)
    \({\left( {\text{sech}\,x} \right)^\prime } \) \(= – \text{sech}\,x\tanh x\)\({\large\int\normalsize} {\text{sech}\,x\tanh xdx} \) \( = – \text{sech}\,x + C\)
    \({\left( {\text{csch}\,x} \right)^\prime } \) \(= – \text{csch}\,x\coth x\)\({\large\int\normalsize} {\text{csch}\,x\coth xdx} \) \(= – \text{csch}\,x + C\)
    \({\left( {\sinh x} \right)^\prime } = \cosh x\)
    \({\large\int\normalsize} {\cosh x dx} = \sinh x + C\)
    \({\left( {\cosh x} \right)^\prime } = \sinh x\)
    \({\large\int\normalsize} {\sinh x dx} = \cosh x + C\)
    \({\left( {\tanh x} \right)^\prime } = {\text{sech}^2}x\)
    \({\large\int\normalsize} {{\text{sech}^2}x dx} = \tanh x + C\)
    \({\left( {\coth x} \right)^\prime } = -{\text{csch}^2}x\)
    \({\large\int\normalsize} {{\text{csch}^2}x dx} = -\coth x + C\)
    \({\left( {\text{sech}\,x} \right)^\prime } \) \(= – \text{sech}\,x\tanh x\)
    \({\large\int\normalsize} {\text{sech}\,x\tanh xdx} \) \( = – \text{sech}\,x + C\)
    \({\left( {\text{csch}\,x} \right)^\prime } \) \(= – \text{csch}\,x\coth x\)
    \({\large\int\normalsize} {\text{csch}\,x\coth xdx} \) \( = – \text{csch}\,x + C\)

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

    \[{u = {e^x},}\;\; \Rightarrow {x = \ln u,\;\;}\kern0pt{dx = \frac{{du}}{u}.}\]


  • 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 the integral \(\int {\large{\frac{{\sinh x}}{{1 + \cosh x}}}\normalsize dx}.\)

    Example 3

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

    Example 4

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

    Example 5

    Evaluate \({\large\int\normalsize} {{{\sinh }^3}xdx}.\)

    Example 6

    Evaluate the integral \({\large\int\normalsize} {x\sinh xdx}.\)

    Example 7

    Evaluate the integral \({\large\int\normalsize} {{e^x}\sinh xdx}.\)

    Example 8

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

    Example 9

    Evaluate the integral \(\int {{e^{-x}}\sinh 2xdx}.\)

    Example 10

    Evaluate the integral \(\large{\int}\normalsize {\large{\frac{{dx}}{{\sinh x}}}\normalsize} .\)

    Example 11

    Find the integral \({\large\int\normalsize} {\large\frac{{dx}}{{1 + \cosh x}}\normalsize}.\)

    Example 12

    Find the integral \({\large\int\normalsize} {\large\frac{{dx}}{{\sinh x + 2\cosh x}}\normalsize}.\)

    Example 13

    Find the integral \(\int {\large{\frac{{dx}}{{\sinh x – \cosh x}}}\normalsize}.\)

    Example 14

    Find the integral \(\int {\large{\frac{{dx}}{{3\sinh x – 5\cosh x}}}\normalsize}.\)

    Example 15

    Evaluate the integral \(\int {\sinh x\cosh \left( { – x} \right)dx}.\)

    Example 16

    Calculate the integral \({\large\int\normalsize} {\sinh 2x\cosh 3xdx}.\)

    Example 17

    Find the integral \(\int {\tanh 2xdx}.\)

    Example 18

    Find the integral \(\int {\coth \large{\frac{x}{3}}\normalsize dx}.\)

    Example 19

    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 the integral \(\int {\large{\frac{{\sinh x}}{{1 + \cosh x}}}\normalsize dx}.\)

    Solution.

    Using the substitution

    \[{u = 1 + \cosh x,\;\;}\kern0pt{du = \sinh xdx,}\]

    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 \({\large\int\normalsize} {{{\sinh }^3}xdx}.\)

    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

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

    Example 6.

    Evaluate the integral \({\large\int\normalsize} {x\sinh xdx}.\)

    Solution.

    We use integration by parts: \({\large\int\normalsize} {udv} \) \(= uv – {\large\int\normalsize} {vdu} .\) Let \(u = x,\) \(dv=\sinh xdx.\) Then, \(du = dx,\) \(v = {\large\int\normalsize} {\sinh xdx} \) \(= \cosh x.\) Hence, the integral is

    \[
    {\int {x\sinh xdx} }
    = {{x\cosh x }-{ \int {\cosh xdx} }}
    = {x\cosh x – \sinh x }+{ C.}
    \]

    Page 1
    Problems 1-6
    Page 2
    Problems 7-19