How to evaluate an indefinite integral

[y_lindsay]

how to evaluate the indefinite integral $$\int \frac{1}{\sqrt{x^2-1}} dx$$

 

[EnumaElish]

Do you have any thoughts of your own?

 

[HallsofIvy]

Is this a test for you or for us?

 

[y_lindsay]

Actually i have solved this problem by using the substitution x=sec u. I'm just thinking if there is any other way to solve this problem?

and another question is how can I delete my post if I find it not worth discussion any more? (I'm new to this forum)

Thanks a lot.

 

[HallsofIvy]

If you look at https://www.physicsforums.com/showthread.php?t=205954 it is exactly the same question. arildno suggested x= cosh(u).

 

[lazypast]

cant solve it using x=sec(theta)as it becomes

$$\int \frac {sec^2 \theta}{tan \theta}d \theta$$

and further simplifyin i get

$$2\int sin^{-1}2\theta$$

 

[rohanprabhu]

cant solve it using x=sec(theta)as it becomes

$$\int \frac {sec^2 \theta}{tan \theta}d \theta$$

and further simplifyin i get

$$2\int sin^{-1}2\theta d \theta$$

so, you are getting:

$$2\int \frac{1}{sin 2\theta} d \theta = 2\int cosec 2\theta$$

and..

$$2\int cosec 2\theta d \theta = 2 \times \frac{1}{2} \times log|tan(\theta)| = log|tan(\theta)| + c$$

 

[lazypast]

yeah that's right. i wouldve attepmted the trig integral but have no idea how to even start them other.

my first guess would be the integral of the cosec expansion, from the maclaurin series

 

[mathwonk]

the way this old game is played is to use trig substitutions and try to show the integral agrees with some elementary combination of trig and exponential functions. but that is just a classical game, with few important consequences.

I think a more intelligent and practical real life solution is yours, using a power series to approximate the value, and similarly that of hundreds of others which do not yield to the tricks of the classical game.

 

[rohanprabhu]

yeah that's right. i wouldve attepmted the trig integral but have no idea how to even start them other.

my first guess would be the integral of the cosec expansion, from the maclaurin series

Here is a small and effective proof for this:

$$\int cosec(\theta) d\theta = \int\frac{cosec(\theta)(cosec(\theta) - cot(\theta))}{cosec(\theta) - cot(\theta)} d\theta$$

If you differentiate the denominator in this question, you get the numerator. So, this results to:

$$log|cosec(\theta) - cot(\theta)|$$

Having done this, you can easily show using trignometry that:

$$cosec(\theta) - cot(\theta) = tan(\frac{\theta}{2})$$

I guess this is what mathwonk referred to as the 'classical' game.. well.. this is a way.. which works for some cases.. in others the power series is the way to go :D

 

[Gib Z]

cant solve it using x=sec(theta)as it becomes

$$\int \frac {sec^2 \theta}{tan \theta}d \theta$$

and further simplifyin i get

$$2\int sin^{-1}2\theta$$

so, you are getting:

$$2\int \frac{1}{sin 2\theta} d \theta = 2\int cosec 2\theta$$

and..

$$2\int cosec 2\theta d \theta = 2 \times \frac{1}{2} \times log|tan(\theta)| = log|tan(\theta)| + c$$

There was no need for ANY of that :( Even if you don't choose the hyperbolic substitution, you should have stopped at $$\int \frac {sec^2 \theta}{tan \theta}d \theta$$. Tell me you see the easier way to do that!

 

[Winzer]

Lol. Gib Z is totally right.
The same answer is obtained with a simple u sub of the tan function, resulting in the same answer.

However I still like rohanprabhu way and the power series way also.
But its a good thing to learn how to get the same answer multiple ways. This of course depends on your patience.

 

[rohanprabhu]

There was no need for ANY of that :( Even if you don't choose the hyperbolic substitution, you should have stopped at $$\int \frac {sec^2 \theta}{tan \theta}d \theta$$. Tell me you see the easier way to do that!

OMFG.. lol :D

I just didn't look at that.. I just looked at $\int \frac{1}{sin(\theta)}$ and continued with it.. I didn't pay attention to $\int \frac {sec^2\theta}{tan \theta}$.. which ofcourse as u said.. is of the form $\int \frac {f(x)}{f'(x)}dx$...

-1

 

[y_lindsay]

hi, lazypast, actually the original problem does not become $$\int\frac{sec^{2}\theta}{tan\theta}d\theta$$by using x=sec(theta), it becomes $$\int sec\theta d\theta$$

however anyway, lazypast's little mistake has led to an inspiring discussion on how to deal with integration problems. and I really appreciate the power series approximation way to do it.

 

[lazypast]

OMFG.. lol :D

I just didn't look at that.. I just looked at $\int \frac{1}{sin(\theta)}$ and continued with it.. I didn't pay attention to $\int \frac {sec^2\theta}{tan \theta}$.. which ofcourse as u said.. is of the form $\int \frac {f(x)}{f'(x)}dx$...

-1

i didnt actually notice $\int \frac {f(x)}{f'(x)}dx$ could be applied to the sec^2/tan till you just said it.

and since i looked in more detail to it, the result is in terms of theta. and the original expression was in terms of x.

as we said x=sec(theta), replacing the answer to in terms of x instead of theta becomes

ln(tan(sec^-1(x)) +c

an edit- just seen $tan(sec^{-1}(x)) = sinx$ ??