How do I integrate this? Help!

Hey! I've been trying to integrate this but I have no clue really on what to do here?

So the question is what steps would you take in integrating '1/sqrt(x^2 - 2)' wrt x.

I'm thinking that you may need to use some trig substitution like letting x equal to something. But I'm not entirely sure how you would work out what trig substitution to use.

What could I do then to come to an answer?

Any help would be appreciated! Thanks!

(edited 5 years ago)

Reply 1

username3249896

12

If you know what the integral of sec x is then a substitution of x = sqrt 2 sec u will help

Reply 2

Yatayyat

OP

14

Original post by BobbJo

If you know what the integral of sec x is then a substitution of x = sqrt 2 sec u will help

Thanks, but how did you know in the first place that you needed to use a trig substitution of sqrt(2) sec(theta)?

Reply 3

username3249896

12

Original post by Yatayyat

Thanks, but how did you know in the first place that you needed to use a trig substitution of sqrt(2) sec(theta)?

It is an integral of a function of the form 1/sqrt(x^2-a^2). This is an integral which can be solved using the substitution x = a sec theta. This is because the argument of the sqrt function resembles the identity sec^2 x -1 = tan^2 x which will greatly simplify things

Reply 4

DFranklin

18

Original post by Yatayyat

Thanks, but how did you know in the first place that you needed to use a trig substitution of sqrt(2) sec(theta)?

You want to choose a substitution so that x^2 - 2 becomes easy to take the square root of (effectively, "something squared").

If you choose x = sqrt(2) sec(theta), then x^2 = 2 sec^2 theta, and then x^2 - 2 is 2(sec^2 theta - 1) which is 2 tan^2 theta, and then the sqrt is sqrt(2) tan theta.

Reply 5

DFranklin

18

Original post by Yatayyat

Thanks, but how did you know in the first place that you needed to use a trig substitution of sqrt(2) sec(theta)?

Perhaps worth noting that this is probably the most difficult/unusual/obscure "standard integral" in Maths (not necessarily in FM), so

(a) it's worth being sure that this is really what you are supposed to be integrating.
(b) don't feel bad if you couldn't see what to do.

Reply 6

Yatayyat

OP

14

Original post by BobbJo

It is an integral of a function of the form 1/sqrt(x^2-a^2). This is an integral which can be solved using the substitution x = a sec theta. This is because the argument of the sqrt function resembles the identity sec^2 x -1 = tan^2 x which will greatly simplify things

Ahh I see what you mean. Is this a rule that can also apply to anything else too? lets say if then denominator was sqrt(x^2 - 3) then then x = sqrt(3) sec theta, right?

Reply 7

username3249896

12

Original post by Yatayyat

Ahh I see what you mean. Is this a rule that can also apply to anything else too? lets say if then denominator was sqrt(x^2 - 3) then then x = sqrt(3) sec theta, right?

Yes this is true

Reply 8

Yatayyat

OP

14

Original post by DFranklin

Perhaps worth noting that this is probably the most difficult/unusual/obscure "standard integral" in Maths (not necessarily in FM), so

(a) it's worth being sure that this is really what you are supposed to be integrating.
(b) don't feel bad if you couldn't see what to do.