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The Fundamental Hyperbolic Integrals

April 6th, 2009 |

Author: Merrin

Example 1. Solve the fundamental hyperbolic integrals

$\displaystyle \int \text{sinh}\, x dx \qquad \displaystyle \int \text{tanh}\, x dx \qquad \displaystyle \int \text{sech}\, xdx$

$\displaystyle \int \text{cosh}\, x dx \qquad \displaystyle \int \text{coth}\, x dx \qquad \displaystyle \int \text{csch}\, xdx$

Solution 1. Solving the fundamental hyperbolic integrals goes similar to solving the fundamental trigonometric integrals. Since we more or less know what to expect, we can solve these integrals relatively quickly. Since the derivative of hyperbolic sine is hyperbolic cosine and vice versa their integrals are simple.

$\int \text{sinh}\, xdx=\text{cosh}\, x+C$

$\int \text{cosh}\, xdx=\text{sinh}\, x+C$

The third and fourth fundamental hyperbolic integrals can be found from a simple \textit{u} substitution.

$\int \text{tanh}\, xdx=\displaystyle \int \cfrac{\text{sinh}\, xdx}{\text{cosh}\, x} =\displaystyle \int \cfrac{du}{u} =\ln \left|\text{cosh}\, x\right| +C$

$\int \text{coth}\, xdx=\displaystyle \int \cfrac{\text{cosh}\, xdx}{\text{sinh}\, x} =\displaystyle \int \cfrac{du}{u} =\ln \left|\text{sinh}\, x\right| +C$

The integrals of hyperbolic secant and hyperbolic cosecant can be found with a slight algebraic manipulation of the integrand.

$\displaystyle \int \text{sech}\, xdx=\displaystyle \int \cfrac{\text{cosh}\, x}{\text{cosh}\, ^{2} x} dx =\displaystyle \int \cfrac{\text{cosh}\, xdx}{1+\text{sinh}\, ^{2} x}$

$I=\displaystyle \int \cfrac{du}{1+u^{2} } =\text{arctan}\,\, (u)+C$

$\int \text{sech}\, xdx=\text{arctan}\,\, (\text{sinh}\, x)+C$

Similarly,

$\displaystyle \int \text{csch}\, xdx=\displaystyle \int \cfrac{\text{sinh}\, xdx}{\text{sinh}\, ^{2} x} =\displaystyle \int \cfrac{\text{sinh}\, xdx}{\text{cosh}\, ^{2} x-1} =-\displaystyle \int \cfrac{du}{1-u^{2} }$