Integration using tables

[stoic sky]

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[calm cove]

Well, the problem shouts substituting r = sec(u)

[twin hatch]

I don't know how to do that

My class hasn't covered that

[calm cove]

Let's see, are you familiar with u-sub? 'Cause this is u-sub

[twin hatch]

yeah ofc

[calm cove]

What kind of u have you substituted before?

[twin hatch]

why sec(u)

[calm cove]

r^2 - 1 simplifies to tan^2(u)

[twin hatch]

ooooh

(x^(2)-1=tan^2(u)?)

[calm cove]

r^2 instead of x^2; which would yield what I have already written

[twin hatch]

ik

ok

well there is another that this can be done?

or is it only trig?

[calm cove]

I am heading off to work, let's see if other helpers have alternative ideas

[twin hatch]

alright

thank you though

[misty fable]

Hm.... We can try the following.
u = √(r^2 - 1), du = rdr/√(r^2 - 1), rdr = udu
(r^2 - 1)^(3/2) dr/r = (r^2 - 1)^(3/2) rdr/r^2 = u^4 du/(u^2 + 1)
This is now a rational function.

[twin hatch]

yep

I am getting it now

but an issue I am having rn

[twin hatch]

the u^2 = v^4 thing doesnt make sense to jme

[misty fable]

Well, clearly, if v = u^(1/2), then v^4 = u^2. Though, this approach is the same as mine, but done in two steps instead of one.

[twin hatch]

oooh

can we just work the u^(3/2)/u+1

without doing u-sub again

[misty fable]

That can't be integrated directly.

[twin hatch]

oh ok

I see

that makes alot of sense now

ty

[misty fable]

Notice that they did u = x^2 - 1 and then v = u^(1/2). That is equivalent to by substitution u = √(r^2 - 1), but just done in two steps.

Can you integrate u^4/(u^2 + 1)?

[twin hatch]

yeah

you need to split into two

[misty fable]

Alright, and how would you do that?

[twin hatch]

it would just be

I think you would have use partial fractions

so it would be 1/(u^2+1) - (u^2-1)?

int(1/u^2+1)+int(u^(2))-int(1)

is this right?

[misty fable]

Yeah, nice! Though, that isn't a result of partial fractions. The approach here is the following:
u^4/(u^2 + 1) = (u^4 - 1 + 1)/(u^2 + 1) = u^2 + 1 + 1/(u^2 + 1)

[twin hatch]

ok ty

[stable tangle]

@twin hatch has given 1 rep to @misty fable

[twin hatch]

I have figured it out

tysm