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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: ${\int {\sinh x\cos xdx} } = {\frac{{\cosh x\cos x + \sinh x \sin x}}{2}.}$