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Calculus

Integration of Functions

Integration of Hyperbolic Functions

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

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 on problem description to see 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