Integrating (x2-5)1/2: Step-by-Step Guide

[gaobo9109]

Homework Statement

How to you integrate (x²-5)^1/2

Homework Equations

The Attempt at a Solution

I have tried substitution, integration by part, and none seems to work. I really don't know where to start.

 

[dextercioby]

make a substitution:

x=\sqrt{5} \cosh t

 

[gaobo9109]

How did you get this substitution? Hyperbolic function is not in my school syllabus. Is there any other form of substitution to to solve this question?

 

[uart]

How did you get this substitution? Hyperbolic function is not in my school syllabus. Is there any other form of substitution to to solve this question?

Well the cosh substitution is the easiest way, but you can also attack it as follows.

Rearrange the integrand as :

I = \int \frac{x^2-5}{\sqrt{x^2-5}}\, dx

Split into two parts :

I = \int \frac{x^2}{\sqrt{x^2-5}} \, dx - \int \frac{5}{\sqrt{x^2-5}} \, dx

You should be able to evaluate the second part by use of "standard integrals" and if you carefully apply integration by parts to the first integral you can reduce it to an expression minus "I" (where "I" is the original integral).

This let's you get it into the form of : "I = expression - I". Then you just solve that algebraically.

I know that's only an outline, but it does work! See how far you can get with it.

 

[dextercioby]

You can use the connection between the hyperbolic and the circular trigonometric functions and make other changes of variable to your integral.

I= \int \sqrt{x^2 -5} \, dx = \int \sqrt{x^2 +(i\sqrt{5})^2} \, dx \, , \, \forall x>\sqrt{5}

Now you can make the substitution

x= i\sqrt{5} \tan t

However, the road is pretty tough. The hyperbolic function shortens a lot of calculations.

 

[Bohrok]

Or you could try u = √(5)secx

 

[uart]

Or you could try u = √(5)secx

Would you care to elaborate on how that works Bohrok?

 

[rahuld.exe]

use x = root5 sectheta...
i've solved it...works like a dream!...

 

[uart]

use x = root5 sectheta...
i've solved it...works like a dream!...

Ok so that substitution leads to :

5 \, \int \frac{\sin^2(\theta)}{\cos^3(\theta)} \, d \theta

Where did you go from there?

 

[rahuld.exe]

yeah now
5<br /> \int tan\theta tan\theta sec\theta d\theta<br />
put t = sec\theta
therefore dt = sec\theta tan\theta d\theta
draw a right angled triangle... using t = sec\theta.. you'll get value of two sides... find the value of the third side by pythagoras theorem... that way now you can find tan\thetaput the value of tan\theta and dt in the integrand... solve !

 

[uart]

yeah now
5<br /> \int tan\theta tan\theta sec\theta d\theta<br />
put t = sec\theta
therefore dt = sec\theta tan\theta d\theta
draw a right angled triangle... using t = sec\theta.. you'll get value of two sides... find the value of the third side by pythagoras theorem... that way now you can find tan\thetaput the value of tan\theta and dt in the integrand... solve !

Is that meant to be a joke? It obviously leads you around in a big circle back to the original integral.

Look there were two perfectly good methods presented to solve this in replies #2 and #4. Enough of the junk replies please.

 

[rahuld.exe]

sorry i had made a mistake... i forgot the root sign... but i think this time i got it correct..
could you please check it for me...
PS: moderators... i m not posting the solution... i am actually checking if what I've done is correct...

http://img831.imageshack.us/i/p1010910lr.jpg/

 

[uart]

sorry i had made a mistake... i forgot the root sign... but i think this time i got it correct..
could you please check it for me...
PS: moderators... i m not posting the solution... i am actually checking if what I've done is correct...

http://img831.imageshack.us/i/p1010910lr.jpg/

Hi rahuld.exe. Yes that still has an error, in the triangle you forgot the sqrt sign.

Where you got \sec(\theta) = t^2 + 1 it should have been \sec(\theta) = \sqrt{(t^2 + 1)}. So you still end up with a difficult integral.

PS. Sorry if I was rude with the previous reply but I wasn't sure whether or not your effort was serious. Now I see you are making a genuine effort, the difficulty is just that the original x=\sqrt{5} \sec(\theta)[/tex] substitution isn&#039;t all that useful and that&#039;s what&#039;s making it hard for you to proceed with that method.<br /> <br /> Take a look at the method I recommended in reply #4, you&#039;ll get it out in a just a few lines! In that method use the following &quot;standard integral&quot;,<br /> <br /> \int \frac{1}{\sqrt{x^2-a^2}} \, dx = \log_e(x + \sqrt{x^2 - a^2 })<br /> <br /> BTW The LHS above is actually inverse_cosh (hence the connection to bigubau&#039;s method). But most standard integral tables give it in the logarithmic form above, so we can keep completely clear of the hyperbolics with this method.

 

[uart]

One last hint re the method of reply #4. Where "integration by parts" is called for use :

\int \frac{x^2}{\sqrt{x^2-5}} \, dx = \int x \, \frac{d}{dx}\left\{\sqrt{x^2-5}\right\} \, dx

 

[rahuld.exe]

hey... i am new to integration so am not that good at it... i got it with your method and hint.. thanks !