# 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 or tap 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

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

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Problems 1-6
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Problems 7-19