Integration sidetracks

[Deleted member 4993]

You are making this waaaay too difficult.

[imath]x = sec( \theta )[/imath]

So [imath]x^2 - 1 = sec^2( \theta ) - 1 = tan^2( \theta )[/imath]

and [imath]dx = sec( \theta ) ~ tan( \theta ) ~d \theta[/imath]

Your integral turns into
[math]\int \dfrac{dx}{(x^2 - 1)^{3/2}} = \int \dfrac{ sec( \theta ) ~ tan( \theta ) ~ d \theta }{(tan^2( \theta ))^{3/2} }[/math]
See what you can do with this.

-Dan

I have been going round in circles on this and any help would be greatly appreciated

Dan, now you are going to make the lad go in circles counter-clockwise and make him cry for his uncle......

Now, at least he will run around in the (+) direction.

 

[Steven G]

Lad, I have been working on this problem for a month and I was hoping for some help. If you can't be bothered then thanks, but no thanks

Dr Khan is not a 'Lad'.
If you equate help with having your problem solved for you, then you are at the wrong forum. Maybe you should try www.haveyourmathhomeworkdoneforyouforfree.com

 

[Deleted member 4993]

Dr Khan is not a 'Lad'.

Steven,

you have to overlook his "post colonial" air of superiority - he has not been taught better.

 

[Shiloh]

Dr Khan is not a 'Lad'.
If you equate help with having your problem solved for you, then you are at the wrong forum. Maybe you should try www.haveyourmathhomeworkdoneforyouforfree.com

Oh, darn! There doesn't appear to be such a site!

 

[Shiloh]

Your a bore, Chester!

 

[topsquark]

Your a bore, Chester!

You're a bore, Chester!

-Dan

 

[Shiloh]

Oh, yes! How embarassing!

 

[Steven G]

To all admins:
I apologize for including a url in my above post. I just remembered that urls are not allowed. I guess it was overlooked since it was a fake url anyways.

 

[borchester]

Steven,

you have to overlook his "post colonial" air of superiority - he has not been taught better.

Subhotosh Khan remains a pompous fool, but that is his problem and I will leave it for him to solve.

That said, as a result of Dr Petersen's and Topsquark's advice I have cracked the problem. So many thanks to them.

 

[topsquark]

Subhotosh Khan remains a pompous fool, but that is his problem and I will leave it for him to solve

You're new. We can forgive you for this.

-Dan

 

[Steven G]

That's very generous of you Dan.
I on the other hand will ask Subhotosh to issue a warning to borchester.

 

[Shiloh]

But he's OUR pompous fool

(No one has called me a pompous fool Why am I being neglected?)

 

[Otis]

urls are not allowed

Is that another new rule?



[imath]\;[/imath]

 

[borchester]

You're new. We can forgive you for this.

-Dan

That bloody question was driving me crazy, so I would forgive you the killing of the first born. Plus, I have a pile of other questions to ask....

But he's OUR pompous fool

(No one has called me a pompous fool Why am I being neglected?)

You must try harder

 

[Steven G]

Is that another new rule?



[imath]\;[/imath]

It's not a new rule. Actually rethinking it, it is not a rule at all. What is a rule is that you can't use a url that goes to a business.