Solving Definite Integral with Inverse Hyperbolic Identities

[2^Oscar]

Hey guys,

I was doing some practice questions and this particular one has me stumped. The topic was on integration with inverse hyperbolic identities, and I was asked to give exact solutions for the following integral:

\int\sqrt{4x^2 -1} dx between \frac{1}{2} and \frac{13}{10}


From looking around on the internet I have found a standard integral that I can use (http://en.wikipedia.org/wiki/List_of_integrals_of_irrational_functions" ) but I would quite like to know the process of deriving this.

I have been fairly confident with these questions but I can see no way of using the basic inverse hyperbolic identities to reach this result and express the definite integral exactly.


Could anyone please lend a hand?


Thanks in advance,
Oscar

 

[HallsofIvy]

Well, this is, after all, in a section about hyperbolic functions! I remember that cosh^2(x)- sinh^2(x)= 1 so that (dividing through by cosh^2(x)) 1- tanh^2(x)= sech^2(x). I would try the substitution 2x= tanh(x).

 

[jbunniii]

Write

x = \frac{1}{2} \cosh y

Then

dx = \frac{1}{2} \sinh y dy

The indefinite integral then becomes

\frac{1}{2} \int \sqrt{\cosh^2 y - 1} \sinh y dy

You should then be able to find the indefinite integral of this if you use

\cosh^2 y - \sinh^2 y = 1

and the definition of \sinh y. After you do that, it's pretty easy to rewrite the answer in terms of x instead of y, and plug in the endpoints in terms of x.

 

[2^Oscar]

Hey,

Thank you for your replies.

Solved the problem now :)

Thanks,
Oscar

 

[arildno]

Well, your integral is therefore:
\frac{1}{2}\int_{0}^{cosh^{-1}(\frac{13}{5})}sinh^{2}ydy
We have the following anti-derivative:
\int\sinh^{2}{y}dy=\frac{1}{2}(\cosh(y)\sinh(y)-y)
Using this, we get:
<br /> \frac{1}{2}\int_{0}^{cosh^{-1}(\frac{13}{5})}sinh^{2}ydy=\frac{1}{4}(\frac{156}{25}-cosh^{-1}(\frac{13}{5})

Since cosh^{-1}(x)=\ln(x+\sqrt{x^{2}-1}), the answer can be simplified to:
\frac{39}{25}-\frac{\ln(5)}{4}

 

[jbunniii]

By the way, if you want to see WHY we have the indicated anti-derivative, then use the definition of sinh:

\int \sinh^2 y dy = \frac{1}{4} \int (e^y - e^{-y})^2 dy

and work from there.

 

[2^Oscar]

thank you again for the speedy replies :)

I got the same answer as you, arildno, however I did it using different method...



Oscar