From this point I'm not sure what to do. I know of the identities:
However, even with these in mind, I've been trying this for ages to no avail. Can someone give me some pointers?
Dec 2nd 2008, 10:20 AM
But this is false !
It's a multiplication, not an addition.
will give you what you wrote, but the multiplication will give :
(you can simplify by 2) :
But I'd suggest you use these formulae :
Divide both numerator and denominator by (Wink)
Dec 2nd 2008, 10:39 AM
Ahh, such as stupid mistake, I've been looking at it so long I missed it. (Giggle)
Dec 2nd 2008, 11:32 AM
Ok, now I'm trying to prove:
Any ideas? I'm not sure if this is even right, it was after an integration of
Dec 2nd 2008, 12:54 PM
This is not true . . .
Dec 2nd 2008, 03:23 PM
Hi Soroban, I suspected as much, since I was getting that same expression in trying to simplify. It must be a problem with my integration then. I used the substitution , and I think my mistake was taking the out of the integral, which can't be done.
Is there some easier substitution I can make?
Dec 2nd 2008, 03:55 PM
Hi, no worries, I've cracked it now...proved that the derivative of sinhx is coshx, therefore the integral of coshx is sinhx. Thanks for the help though!