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Integration

I'm not which substitution to use to get from 1 / z*(z^-1)^.5

To arcsin(1/z)

Thanks in advance
(edited 11 years ago)
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Reply 1
Avatar for Notnek
Notnek
21
I'm not sure what you mean when you say, "the second going into ln".

Could you clarify which line/part you are stuck on?
Reply 2
Avatar for james303
james303
OP
7
Original post by notnek
I'm not sure what you mean when you say, "the second going into ln".

Could you clarify which line/part you are stuck on?


Integral (z / (z^2 -1)^2))
Reply 3
Avatar for Notnek
Notnek
21
Original post by james303
Integral (z / (z^2 -1)^2))

I assume that's a typo and you mean a power of 1/2 instead of 2.

One way to do this is to notice that z is constant away from the derivative of z2−1z^2-1, so you can use recognition.

Or you could use the substitution u2=z2−1u^2=z^2-1
Reply 4
Avatar for james303
james303
OP
7
Original post by notnek
I assume that's a typo and you mean a power of 1/2 instead of 2.

One way to do this is to notice that z is constant away from the derivative of z2−1z^2-1, so you can use recognition.

Or you could use the substitution u2=z2−1u^2=z^2-1


Cheers :colondollar:
Reply 5
Avatar for james303
james303
OP
7
One more problem....

I'm not which substitution to use to get from 1 / z*(z^-1)^.5

To arcsin(1/z)



Thanks
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Reply 6
Avatar for DFranklin
DFranklin
18
How is this different from your original post? :confused:
Reply 7
Avatar for Notnek
Notnek
21
Original post by james303
One more problem....

I'm not which substitution to use to get from 1 / z*(z^-1)^.5

To arcsin(1/z)

Thanks

Try the substitution z=sec⁡xz=\sec x. You may have to a bit of work at the end to get it in the required form.

Original post by DFranklin
How is this different from your original post? :confused:

He edited his original post after posting this new question. I'm not sure why...

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