calculuspowerup.com

Integration by Hyperbolic Substitution

April 6th, 2009 |

Author: Merrin

Trigonometric substitutions are not the only game in town. Hyperbolic substitutions are also useful. Recall when we figured out which substitutions eliminated the radical from the form we was using trigonometric identities. Now we can do the same for hyperbolic functions

$\text{cosh}\, ^{2} u-\text{sinh}\,^{2} u=1$

This suggests for the form

$(x^{2} -a^{2} )^{n+1/2}$

you can do a substitution of x= a cosh(u) resulting in a transformed form of

$(a^{2} \text{cosh}\,^{2} u-a^{2} )^{n+1/2} =(a\text{sinh}\,u)^{2n+1}$

Dividing the main hyperbolic identity by cosh u twice makes the new identity

$1-\text{tanh}\,^{2} u = \text{sech}\,^{2} u$

This suggests for the form

$(a^{2} -x^{2} )^{n+1/2}$

a substitution of x=a tanhu results in a transformed form of

$(a^{2} -a^{2} \text{tanh}\,^{2} u)^{n+1/2} =(a\text{sech}\, u)^{2n+1}$

There is one last form to do for hyperbolic substitution which is the form of

$(a^{2} +x^{2} )^{n+1/2}$

where we choose x=a sinhu which leads to the form

$(a^{2} +a^{2} \text{sinh}\,^{2} u)^{n+1/2} =(a\text{cosh}\,u)^{2n+1}$

There is no obvious reason why you should choose trigonometric substitution over hyperbolic substitution other than you might be more proficient with one type of integrals over another.

Table of possible trigonometric and hyperbolic substitutions

$\sqrt{a^{2} -x^{2} } \qquad x=a\sin \theta \qquad dx=a\cos \theta d\theta$

$\sqrt{a^{2} -x^{2} } \qquad x=a\tanh u \qquad dx=a \text{sech}^2\,u du$

$\sqrt{a^{2} + x^{2} } \qquad x=a\tan \theta \qquad dx=a\,\sec^2 \theta d\theta$

$\sqrt{a^{2} + x^{2} } \qquad x=a\sinh u \qquad dx=a\cosh u du$

$\sqrt{x^{2} - a^{2} } \qquad x=a\cosh u \qquad dx=a\sinh u du$

$\sqrt{x^{2} - a^{2} } \qquad x=a\sec \theta \qquad dx=a\sec \theta \tan \theta d\theta$

I would suggest for the form

$\sqrt{a^{2} -x^{2} }$

to use the trigonometric substitution but for the other two forms use the hyperbolic substitutions. The integration may be easier via this route.

Intermediate calculations occur for the hyperbolic substitutions so you need to know hyperbolic identities and what I call the “hyperbolic reference triangle.” You have almost all the same rules except that the equivalent of the hypotenuse is calculated differently.

$h^{2} =x^{2} -y^{2}$

hyperref

Reading off the parameter from the hyperbolic inverse functions is no different than for trigonometric reference triangles once you have set up the hyperbolic reference triangle.

$u={\text{arcsinh}\,(y/\sqrt{x^{2} -y^{2} } )=\text{arccosh}\,(x/\sqrt{x^{2} -y^{2} } )=\text{arctanh}\,(y/x)} \\ u= { \text{arccsch}\,(\sqrt{x^{2} -y^{2} } /y)=\text{arcsech}\,(\sqrt{x^{2} -y^{2} } /x)=\text{arccoth}\,(x/y)}$

Whichever choice you make trigonometric or hyperbolic you are well equipped to solve the integral and do any intermediate computations. For the three main substituions the hyperbolic functions are calculated from the sides of the triangle just like you would for trigonometric functions. For example for x = a sinh(u) then tanh(u) = opposite over adjacent equals x/sqrt(a^2+x^2)

hyperref21

For the following examples I am going to use the parameter b instead of a because something like atan(x) could be confused with arctan(x). I never use the a to symbolize the arc but some people do use bad notations.

Example 1. Integrate

$\displaystyle \int (b^{2}-x^{2} )^{3/2} dx$

Solution 1. Although this section is on hyperbolic substitution I still suggest to do the trigonometric substitution for this type because it is simpler. The suggested trigonometric substitution is

$x= b \sin \theta$

We have then that

$dx=b\cos\theta d\theta$

Putting all the pieces in the integral we get

$\int (b^2-b^2\sin^2 \theta )^{3/2} b\cos \theta d\theta }={b^4\displaystyle \int \cos^{4} \theta d\theta$

We have reduce the problem to a situation we already know which is a reduction formula. The reduction formula are useful because the frequency of the sines and cosine in the antiderivative are just one. If you integrate by a different method you have to convert everything to be compatible with the back transformation of the reference triangle.

$\displaystyle \int (\cos x)^{n} dx =+\cfrac{(\sin x)(\cos x)^{n-1} }{n} +\cfrac{n-1}{n}\displaystyle \int (\cos x)^{n-2} dx$
$\displaystyle \int \cos^4\theta d\theta=\frac{\sin\theta(\cos\theta)^3}{4}+\cfrac{3}{4}(x/2+\sin(2\theta)/4)+C$

$I = \cfrac{3}{8} \text{arcsin}\,(x/b)+\cfrac{x}{4b^4}\left( b^2-x^2 \right)^{3/2} + \cfrac{3x}{8b^2} \sqrt{b^2-x^2}+C$

Example 2. Integrate

$I=\displaystyle \int \sqrt{b^{2} +x^{2} } dx$

Solution 2. The hyperbolic substitution is simpler in this case. We have that

$x=b \sinh u$

$dx=b\cosh udu$

When we make the substitution, we have

$I=\displaystyle \int b^{2} \sqrt{1+\sinh^2 u} \cosh udu$

$I = b^{2} \displaystyle \int \cosh^{2} udu =b^2 \displaystyle \int \cfrac{1}{2} \left( 1+\cosh 2u \right)du \\I=\cfrac{b^{2} }{2} (u+\cfrac{1}{2} \sinh 2u)=\cfrac{b^{2} }{2} (u+\sinh u\cosh u)+C \\I=\cfrac{b^{2} }{2} \text{arcsinh}\, (x/b)+\cfrac{x}{2} \sqrt{b^2+x^2} +C$

We used our hyperbolic reference triangle on the u term and

$\sinh u\cosh u$

Example 3. Integrate

$\displaystyle \int \cfrac{dx}{\sqrt{b^{2} +x^{2} } }$

Solution 3. Normally you would do a

$x=b\,\tan \theta$

here but instead we try the improved recommendation of a hyperbolic substitution

$x=b\,\cosh u$

$\int \cfrac{dx}{\sqrt{b^2 +x^2 } } =\displaystyle \int \cfrac{dx}{\sqrt{b^2 +b^2 \sinh^2 u} } =\displaystyle \int \cfrac{b\cosh udu}{b\cosh u} =u+C$
${I=\text{arccosh}\,(x/b)+C}$

Otherwise this integral can also be done using a trigonometric substitution.

${x=b\tan \theta }$
$\int \cfrac{b\sec^2 \theta d\theta }{b\sec \theta } =\displaystyle \int \sec \theta d\theta =\ln \left|\sec \theta +\tan \theta \right|+C$
${I=\ln \left|\cfrac{x}{b} +\cfrac{x}{\sqrt{b^{2} +x^{2} } } \right|+C}$

The algebraic expression for

$\text{arccosh}\,(x/b)$

can be derived as an inverse and matches this integral.

The question of integration of certain irrational and rational forms can be aided by the calculation of trigonometric and hyperbolic substitutions. Intermediate calculations can be done using reference triangles either trigonometric or hyperbolic.

Back to Home:

Back to Index:

Previous Topic: Integration by Trigonometric Substitution

Next Topic: Introduction to Differential Equations

Bookmark It