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Intemegration of Hyperbolicss

how do you

Int (1/sinhu) du

for relevance sake it started as Int (1/((x^2)-2x)) dx and i whacked in coshu = x-1

cheers

el stevo
Reply 1
Avatar for kikzen
kikzen
14
um

1/(x^2 - 2x) = 1/x(x-2)

partial fractions
Reply 2
Avatar for El Stevo
El Stevo
OP
11
i'd do that, but we were specifically told to use substitution coshu=x-1
Reply 3
Avatar for Ben.S.
Ben.S.
11
El Stevo
how do you

Int (1/sinhu) du

for relevance sake it started as Int (1/((x^2)-2x)) dx and i whacked in coshu = x-1

cheers

el stevo



Int[cosech(x)dx] = Int[(cosech(x)^2 + coth(x).cosech(x))/(cosech(x) + coth(x))dx]

= - ln |cosech(x) + coth (x)| + constant

I have missed out some steps at the beginning, but I'll clarify if you need it.

Ben
Reply 4
Avatar for El Stevo
El Stevo
OP
11
i got ya, the integral of cosech is the same as the integral of cosec, written of course with hyperbolic functions...

*quickly fiddles*

i make that equivalent to ln|tanh(x/2)|+c

cheers ben

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