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

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:

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

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

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Solved Problems

Click a problem to see the solution.

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

\[
{\int {\frac{{\cosh x}}{{2 + 3\sinh x}}dx} }
= {\int {\frac{{\frac{{du}}{3}}}{u}} }
= {\frac{1}{3}\int {\frac{{du}}{u}} }
= {\frac{1}{3}\ln \left| u \right| + C }
= {\frac{1}{3}\ln \left| {2 + 3\sinh x} \right| }+{ C.}
\]

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

\[{I = \ln \left( {1 + \cosh x} \right) + C.}\]

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

\[{\int {{{\sinh }^2}xdx} }={ \frac{1}{2}\int {\left( {\cosh 2x – 1} \right)dx} }={ \frac{1}{2}\left( {\frac{{\sinh 2x}}{2} – x} \right) + C }={ \frac{1}{4}\sinh 2x – \frac{x}{2} + C.}\]

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:

\[{\int {{{\cosh }^2}xdx} }={ \frac{1}{2}\int {\left( {\cosh 2x + 1} \right)dx} }={ \frac{1}{2}\left( {\frac{{\sinh 2x}}{2} + x} \right) + C }={ \frac{1}{4}\sinh 2x + \frac{x}{2} + C.}\]

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

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

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

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