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P5 integration Q watch

1. Little bit of trouble with this one. Nearly there though

Intergrate with respect to x: (Q11 P5 ex 2A)

coth^2 (x/2)

This is what I have got:

=> INT cosech^2 (x/2) +1 dx
= INT [sinh (x/2)]^-2 +1 dx
= -(sinh (x/2))^-1 / (0.5cosh(x/2) + x + C <= this is where I think I have gone wrong

= -2/[cosh (x/2)sinh (x/2)] + x + C

This cosh (x/2) is supposed to be on the top of the fraction so it simplifies to coth. cant see how to get it up there though :'(

Also, another one Q 26 in P5 ex 3A I get the final answer as 0.5 arctan(sinh 2x) + C whilst the answer has it as arctan e ^2x. Are these the same?
2. What board is it from? I'm doing edexcel and we are supposed to know that

d(coth(x))/dx = -cosech²x (it's in FB but as trig instead of hyperbolic so can just look at that to help remember)

So from the second line on your working

∫(cosech²(x/2) + 1)dx
= -2coth(x/2) + x
3. Ahhhhhh, I saw that cot x => -cosec^2 x but not coth . Are all the trig ones the same as hyperbolic ones?

Also, tried it a different way and reduced to:

=> 2 INT 1/(coshx - 1) dx Can anyone solve that one? Do you have to use the t substitution? t = tanh x/2? If so, are the sinh and cosh functions the equilivilent of the trig functions?

e.g. sin A = 2t/1+t^2 ==> sinh A = 2t/1+t^2?
4. (Original post by Womble548)
Ahhhhhh, I saw that cot x => -cosec^2 x but not coth . Are all the trig ones the same as hyperbolic ones?

Also, tried it a different way and reduced to:

=> 2 INT 1/(coshx - 1) dx Can anyone solve that one? Do you have to use the t substitution? t = tanh x/2? If so, are the sinh and cosh functions the equilivilent of the trig functions?

e.g. sin A = 2t/1+t^2 ==> sinh A = 2t/1+t^2?
i'm confused about that so when do you apply osbournes rule?

only in identities? and not in calculus?
5. (Original post by Phil23)
i'm confused about that so when do you apply osbournes rule?

only in identities? and not in calculus?
Dunno?!
6. I don't think they are nescesarily going to the the same as trig in calculus and I think osbournes rule is only for identities.

I just use the fact they are similar to help remember, and remember which ones arent the same.

d(sech(x))/dx = -tanh(x)sech(x)

and

d(sec(x))/dx = tan(x)sec(x)

So they arent all the same, and that one doesnt seem to follow osbournes rule?

I dunno whether there is a rule for the calculus ones. Hopefully someone who knows what they're on about will help us
7. (Original post by Womble548)
Dunno?!
Osborne's rule is only valid for identities n NOT for calculus!

there isnt any hard n fast rule for calculus, but more often than not they r comparable

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