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:
$$\frac{1}{2}\int_{0}^{cosh^{-1}(\frac{13}{5})}sinh^{2}ydy=\frac{1}{4}(\frac{156}{25}-cosh^{-1}(\frac{13}{5})$$

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