Intergration Of Hyperbolic Functions

Integration of hyperbolic functions

Since hyperbolic functions are expressed in terms of $e^x$ and $e^{-x}$, the rules for integrating hyperbolic functions can be deduced.

Integrals of common hyperbolic functions

The following are the integrals of common hyperbolic functions:

$\int \sinh x \, dx = \cosh x + c$

$\int \cosh x \, dx = \sinh x + c$

$\int \tanh x \, dx = \ln(\cosh x) + c$

$\int \text{cosech}^2 x \, dx = -\coth x + c$

$\int \text{sech}\,x \tanh x \, dx = -\text{sech}\,x + c$

$\int \text{sech}^2 x \, dx = \tanh x + c$

Integrals of the form $\int \frac{1}{\sqrt{a^2 + x^2}} dx$

Integrals of the form $\int \frac{1}{\sqrt{a^2 + x^2}} dx$ can be solved using hyperbolic substitution. The result is:

$\int \frac{1}{\sqrt{a^2 + x^2}} dx = \sinh^{-1}\left(\frac{x}{a}\right) + c = \ln\left(x + \sqrt{x^2 + a^2}\right) + c$

where $a > 0$.

Integrals of the form $\int \frac{1}{\sqrt{x^2 - a^2}} dx$

Integrals of the form $\int \frac{1}{\sqrt{x^2 - a^2}} dx$ can be solved using hyperbolic substitution involving the hyperbolic cosine function. The result is:

$\int \frac{1}{\sqrt{x^2 - a^2}} dx = \cosh^{-1}\left(\frac{x}{a}\right) + c = \ln\left(x + \sqrt{x^2 - a^2}\right) + c$

where $x \ge a > 0$.

Example 1

Find $\int \sinh^2 x \, dx$.

Solution:

Using the identity $\cosh 2x = \cosh^2 x + \sinh^2 x = 1 + 2\sinh^2 x$, we have $\sinh^2 x = \frac{\cosh 2x - 1}{2}$.

$\int \sinh^2 x \, dx = \int \frac{\cosh 2x - 1}{2} \, dx = \frac{1}{2}\int (\cosh 2x - 1) \, dx = \frac{1}{2}\left(\frac{\sinh 2x}{2} - x\right) + c = \frac{\sinh 2x}{4} - \frac{x}{2} + c$

Example 2

Find $\int \text{sech}\,x \, dx$.

Solution:

$\int \text{sech}\,x \, dx = \int \frac{2}{e^x + e^{-x}} \, dx = \int \frac{2e^x}{e^{2x} + 1} \, dx$

Let $u = e^x$, then $du = e^x dx$.

$\int \frac{2}{u^2 + 1} du = 2\arctan(u) + c = 2\arctan(e^x) + c$

Example 3

Evaluate $\int \frac{1}{\sqrt{x^2 + 2x + 2}} dx$.

Solution:

$\int \frac{1}{\sqrt{x^2 + 2x + 2}} dx = \int \frac{1}{\sqrt{(x+1)^2 + 1}} dx$

Let $u = x + 1$, then $du = dx$.

$\int \frac{1}{\sqrt{u^2 + 1}} du = \sinh^{-1}(u) + c = \sinh^{-1}(x+1) + c = \ln(x+1+\sqrt{(x+1)^2+1})+c = \ln(x+1+\sqrt{x^2+2x+2})+c$

Example 4

Find $\int \frac{1}{\sqrt{x^2 - 6x + 7}} dx$.

Solution:

$\int \frac{1}{\sqrt{x^2 - 6x + 7}} dx = \int \frac{1}{\sqrt{(x-3)^2 - 2}} dx$

Let $x-3 = \sqrt{2}\cosh t$, so $dx = \sqrt{2}\sinh t \, dt$.

$\int \frac{\sqrt{2}\sinh t}{\sqrt{2\cosh^2 t - 2}} dt = \int \frac{\sqrt{2}\sinh t}{\sqrt{2}\sinh t} dt = \int dt = t + c$

Since $x-3 = \sqrt{2}\cosh t$, $\cosh t = \frac{x-3}{\sqrt{2}}$, so $t = \cosh^{-1}\left(\frac{x-3}{\sqrt{2}}\right)$.

Therefore, $\int \frac{1}{\sqrt{x^2 - 6x + 7}} dx = \cosh^{-1}\left(\frac{x-3}{\sqrt{2}}\right) + c = \ln(x-3+\sqrt{x^2-6x+7})+c$

Example 5

Evaluate $\int_0^2 \frac{1}{\sqrt{x^2 + 4}} dx$.

Solution:

Let $x = 2\sinh t$, so $dx = 2\cosh t \, dt$. When $x=0$, $t=0$. When $x=2$, $2 = 2\sinh t$ so $\sinh t = 1$ and $t = \sinh^{-1}(1)$.

$\int_0^{\sinh^{-1}(1)} \frac{2\cosh t}{\sqrt{4\sinh^2 t + 4}} dt = \int_0^{\sinh^{-1}(1)} \frac{2\cosh t}{2\cosh t} dt = \int_0^{\sinh^{-1}(1)} dt = [t]_0^{\sinh^{-1}(1)} = \sinh^{-1}(1) - 0 = \ln(1+\sqrt{2}) \approx 0.8814$

Example 6

Find $\int \frac{1}{x\sqrt{x^2-1}}dx$ where $x\ge1$ in logarithmic form.

Let $x=\cosh t$, then $dx=\sinh t \, dt$ and $\sqrt{x^2-1}=\sqrt{\cosh^2 t -1}=\sinh t$.

$\int \frac{\sinh t \, dt}{\cosh t \sinh t} = \int \frac{dt}{\cosh t} = \int \text{sech } t \, dt = 2\arctan(e^t) + c$

Since $x = \cosh t$, $t = \cosh^{-1} x$.

The integral becomes $2\arctan(e^{\cosh^{-1} x}) + c$.

Or, using the other form of the integral of sech:

$\int \text{sech } t \, dt = \ln|\tanh t + \text{sech } t| + c$

Since $x=\cosh t$, $\tanh t = \frac{\sqrt{x^2-1}}{x}$ and $\text{sech } t = \frac{1}{x}$.

$\int \frac{1}{x\sqrt{x^2-1}}dx = \ln\left|\frac{\sqrt{x^2-1}+1}{x}\right|+c$