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Yet another integration problem

username5737602

This really damages my pride to say announce this but I am struggling on yet another integration problem. The intergral of 1/(x^2 -2)^(1/2).

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username3477548

Original post by tej3141

This really damages my pride to say announce this but I am struggling on yet another integration problem. The intergral of 1/(x^2 -2)^(1/2).

You could simply simplify that using indices rules and then go from there

(edited 2 years ago)

username5737602

OP

Original post by Qxi.xli

You could simply simplify that using indices rules and then go from there

Wdym?

username3477548

Original post by tej3141

Wdym?

Just rewrite as powers of x?

Study Forum Helper

If you've done hyperbolic functions, I would note the identity

\cosh^2 x - 1 = \sinh^2 x

. Otherwise, play with

\sec^2 \theta - 1 = \tan^2 \theta

.

It's not really an obvious one at all if you haven't seen it before.

(edited 2 years ago)

probably a sec() type substitution, so use the
tan^2 = sec^2 - 1
identity.

username5737602

OP

Original post by _gcx

If you've done hyperbolic functions, I would note the identity

\cosh^2 x - 1 = \sinh^2 x

. Otherwise, play with

\sec^2 \theta - 1 = \tan^2 \theta

.

It's not really an obvious one at all if you haven't seen it before.

I have not reached that chapter yet. Let me try with tanx and secx

username3477548

I can't lie I'm confused why is everyone overcomplicating it.
Whatever ig lol

Study Forum Helper

Original post by tej3141

I have not reached that chapter yet. Let me try with tanx and secx

it isn't covered in a-level normal maths, only further. So you probably want to use sec and tan.

username5737602

OP

Original post by Qxi.xli

I can't lie I'm confused why is everyone overcomplicating it.
Whatever ig lol

Because this is the question

username3477548

Original post by ghostwalker

If you think of the form:

\dfrac{1}{\sqrt{x^2-a^2}}

- it's a standard integral.

That's what I thought too

Original post by tej3141

Because this is the question

Are they asking you to find what the grey boxes are or something :confused:

:confused:

username5737602

OP

I haven't came across that from yet. Only 1/(a^2 +x^2) and 1/(a^2 - x^2)^1/2. And I haven't done hyperbolics yet.

Study Forum Helper

Original post by Qxi.xli

That's what I thought too

Are they asking you to find what the grey boxes are or something :confused:

:confused:

they're typing in desmos i think, since its graphing it only lets you type definite integrals. q probs doesn't have limits

(edited 2 years ago)

Original post by Qxi.xli

That's what I thought too

Are they asking you to find what the grey boxes are or something :confused:

:confused:

Rewriting using index laws doesn't work, because of the - 2

username5737602

OP

Original post by _gcx

it isn't covered in a-level normal maths, only further. So you probably want to use sec and tan.

I do further maths too, but I just started core pure 2 a week ago, so I won't reach hyperbolic for like 1/2 more weeks

username5737602

OP

Original post by mqb2766

probably a sec() type substitution, so use the
tan^2 = sec^2 - 1
identity.

So what's the exact substitution I should use?

Original post by tej3141

So what's the exact substitution I should use?

Why not try first? You want to match
x^2 - 2
to
sec^2 - 1
to root tan^2(), so have a think about what x = or u = you could try?

Note its similar to the sqrt(1-x^2) circle integrals based on sin^2 and cos^2.

username3477548

Original post by _gcx

they're typing in desmos i think, since its graphing it only lets you type definite integrals. q probs doesn't have limits

Original post by Sinnoh

Rewriting using index laws doesn't work, because of the - 2

Oh lol ty

username5737602

OP

Thanks for the help everyone. I have finally done it

(edited 2 years ago)

username5737602

OP

Original post by mqb2766

Why not try first? You want to match
x^2 - 2
to
sec^2 - 1
to root tan^2(), so have a think about what x = or u = you could try?

Note its similar to the sqrt(1-x^2) circle integrals based on sin^2 and cos^2.

Thanks. I did it with the substitution 2^1/2(sec(u)). What features of this integral made you realise a sec substation was necessary?

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