Math24.net
Calculus
Home
Calculus
Limits and Continuity
Differentiation
Applications of Derivative
Integration
Sequences and Series
Double Integrals
Triple Integrals
Line Integrals
Surface Integrals
Fourier Series
Differential Equations
1st Order Equations
2nd Order Equations
Nth Order Equations
Systems of Equations
Formulas and Tables
   Integration of Hyperbolic Functions
The \(6\) basic hyperbolic functions are defined by


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

There are the following differentiation and integration formulas for hyperbolic functions:


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

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 \(u = {e^x},\) \(x = \ln u,\) \(dx = {\large\frac{{du}}{u}\normalsize}.\)

   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 \[ {I = \int {\left( {{\cosh^2}x - 1} \right)\sinh xdx} } = {\int {\left( {{u^2} - 1} \right)du} } = {\frac{{{u^3}}}{3} - u + C } = {\frac{{{{\cosh }^3}x}}{3} - \cosh x + C.} \]
   Example 3
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 \[ {\int {x\sinh xdx} } = {x\cosh x - \int {\cosh xdx} } = {x\cosh x - \sinh x + C.} \]
   Example 4
Evaluate the integral \({\large\int\normalsize} {{e^x}\sinh xdx} .\)

Solution.
Since \(\sinh x = {\large\frac{{{e^x} - {e^{ - x}}}}{2}\normalsize},\) 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 5
Find the integral \({\large\int\normalsize} {\large\frac{{dx}}{{1 + \cosh x}}\normalsize}.\)

Solution.
By definition of the hyperbolic cosine, \(\cosh x = {\large\frac{{{e^x} + {e^{ - x}}}}{2}\normalsize}.\) Hence, the integral is equal \[ {\int {\frac{{dx}}{{1 + \cosh x}}} } = {\int {\frac{{dx}}{{1 + \frac{{{e^x} + {e^{ - x}}}}{2}}}} } = {\int {\frac{{2dx}}{{2 + {e^x} + {e^{ - x}}}}} } = {2\int {\frac{{{e^x}dx}}{{2{e^x} + {e^{2x}} + 1}}} } = {2\int {\frac{{d\left( {{e^x} + 1} \right)}}{{{{\left( {{e^x} + 1} \right)}^2}}}} } = { - \frac{2}{{{e^x} + 1}} + C.} \]
   Example 6
Find the integral \({\large\int\normalsize} {\large\frac{{dx}}{{\sinh x + 2\cosh x}}\normalsize} .\)

Solution.
By definition, \(\sinh x = {\large\frac{{{e^x} - {e^{ - x}}}}{2}\normalsize}\) and \(\cosh x = {\large\frac{{{e^x} + {e^{ - x}}}}{2}\normalsize}.\) Hence, \[ {I = \int {\frac{{dx}}{{\sinh x + 2\cosh x}}} } = {\int {\frac{{dx}}{{\frac{{{e^x} - {e^{ - x}}}}{2} + 2 \cdot \frac{{{e^x} + {e^{ - x}}}}{2}}}} } = {\int {\frac{{2dx}}{{{e^x} - {e^{ - x}} + 2{e^x} - 2{e^{ - x}}}}} } = {2\int {\frac{{dx}}{{3{e^x} + {e^{ - x}}}}} } = {2\int {\frac{{{e^x}dx}}{{3{e^{2x}} + 1}}} .} \] We make the substitution: \(u = {e^x},\) \(du = {e^x}dx\) and calculate the initial integral: \[ {I = 2\int {\frac{{{e^x}dx}}{{3{e^{2x}} + 1}}} } = {2\int {\frac{{du}}{{3{u^2} + 1}}} } = {\frac{2}{3}\int {\frac{{du}}{{{u^2} + \frac{1}{3}}}} } = {\frac{2}{3}\int {\frac{{du}}{{{u^2} + {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}} } = {\frac{2}{3} \cdot \frac{{\sqrt 3 }}{1}\arctan \frac{u}{{\frac{1}{{\sqrt 3 }}}} + C } = {\frac{2}{{\sqrt 3 }}\arctan \left( {\sqrt 3 u} \right) + C } = {\frac{2}{{\sqrt 3 }}\arctan \left( {\sqrt 3 {e^x}} \right) + C.} \]
   Example 7
Calculate the integral \({\large\int\normalsize} {\sinh 2x\cosh 3xdx} .\)

Solution.
Applying the formulas \(\sinh x = \large\frac{{{e^x} - {e^{ - x}}}}{2}\normalsize\) and \(\cosh x = \large\frac{{{e^x} + {e^{ - x}}}}{2}\normalsize,\) we get \[ {\int {\sinh 2x\cosh 3xdx} } = {\int {\frac{{{e^{2x}} - {e^{ - 2x}}}}{2} \cdot \frac{{{e^{3x}} + {e^{ - 3x}}}}{2}dx} } = {\frac{1}{4}\int {\left( {{e^{2x}} - {e^{ - 2x}}} \right)\left( {{e^{3x}} + {e^{ - 3x}}} \right)dx} } = {\frac{1}{4}\int {\left( {{e^{2x + 3x}} - {e^{ - 2x + 3x}} + {e^{2x - 3x}} - {e^{ - 2x - 3x}}} \right)dx} } = {\frac{1}{4}\int {\left( {{e^{5x}} - {e^x} + {e^{ - x}} - {e^{ - 5x}}} \right)dx} } = {\frac{1}{4}\left( {\frac{{{e^{5x}}}}{5} - {e^x} - {e^{ - x}} + \frac{{{e^{ - 5x}}}}{5}} \right) + C } = {\frac{1}{{10}} \cdot \frac{{{e^{5x}} + {e^{ - 5x}}}}{2} - \frac{1}{2} \cdot \frac{{{e^x} + {e^{ - x}}}}{2} + C } = {\frac{{\cosh 5x}}{{10}} - \frac{{\cosh x}}{2} + C.} \]
   Example 8
Evaluate the integral \({\large\int\normalsize} {\sinh x\cos xdx}.\)

Solution.
We use integration by parts. Let \[ {u = \cos x,}\;\; {dv = \sinh xdx,}\;\; {\Rightarrow du = - \sin xdx,}\;\; {v = \int {\sinh xdx} = \cosh x.} \] Then \[ {\int {\sinh x\cos xdx} } = {\cosh x\cos x - \int {\cosh x\left( { - \sin x} \right)dx} } = {\cosh x\cos x + \int {\cosh x\sin xdx}.} \] Apply integration by parts again to the latter integral. Let \[ {u = \sin x,}\;\; {dv = \cosh xdx,}\;\; {\Rightarrow du = \cos xdx,}\;\; {v = \int {\cosh xdx} = \sinh x.} \] Then we have \[ {\int {\sinh x\cos xdx} } = {\cosh x\cos x + \int {\cosh x\sin xdx} } = {\cosh x\cos x + \left( {\sin x\sinh x - \int {\sinh x\cos xdx} } \right).} \] Solving this equation for \({\large\int\normalsize} {\sinh x\cos xdx},\) we obtain the complete answer: \[ {\int {\sinh x\cos xdx} } = {\frac{{\cosh x\cos x + \sinh x \sin x}}{2}.} \]

All Rights Reserved © www.math24.net, 2010-2016   info@math24.net