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I'm unsure of how to integrate this...

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Check out post #16 if ur confused.
wacabac
use x=cos t


Okay, but x=cos t surely only works for when x is between 1 and -1? Right?
Reply 22
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mikeyT
2
W@t3R
No :redface: .... I dunno what's wrong with me.... I haven't slept in 24 hrs so i'm quite tired... it seems like you just re-iterated what's been written in the answer :confused:

I suggest that you sleep on it.
But otherwise y/√y =√y
Look at my post! Post #16! Me! Me!

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BovineBeast
Okay, but x=cos t surely only works for when x is between 1 and -1? Right?


Yes.
Reply 24
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W@t3R
OP
m277
I suggest that you sleep on it.
But otherwise y/√y =√y

I've not got time to sleep atm.... but OMG, YES, i'm so silly!

And thanks wacabac! ok.... time to get redbull - nated again!
wacabac

Yes.


So if you have, say, ∫√(1-x²)dx with limits, I don't know, -2 and 5, you'd have to add three seperate integrals - ie. substituting cosh t, -cosh t, and cos t?
Reply 26
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mikeyT
2
BovineBeast
So if you have, say, ∫√(1-x²)dx with limits, I don't know, -2 and 5, you'd have to add three seperate integrals - ie. substituting cosh t, -cosh t, and cos t?

I'm sure that you just need the one sub x=cosht
m277
I'm sure that you just need the one sub x=cosht


But cosh t never goes below 1, so it's only valid for those values of x greater than or equal to 1. Unless there's something I've missed.
Reply 28
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mikeyT
2
BovineBeast
But cosh t never goes below 1, so it's only valid for those values of x greater than or equal to 1. Unless there's something I've missed.

Just noticed something:
√(1-4)=√-3
√(1-25)=√-24
This integral isn't valid for the reals.

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It an imaginary integrand outside x=-1 to 1
m277
Just noticed something:
√(1-4)=√-3
√(1-25)=√-24
This integral isn't valid for the reals.


Oh, yes, good point. So substituting cos t is perfectly sensible, if you're restricted to reals. My mind is now at rest.

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