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# Integration question involving hyperbolic trig functions... Watch

1. Problem solved, no more help needed.

Find the area enclosed by y=sinh(3x), y=0 and x=ln3

i'm fairly new to hyperbolic trig functions & im stuck on how to do this question, though, even if it wasn't hyperbolic trig i dont think i'd know what this question was on about , any help in getting started with it is appreciated & rep available still
2. Think about what you need to do, you have a graph y=sinh(3x) and you need to find the area between two points, x=0 and x=ln(3). How would you usually calculate areas between points?
3. (Original post by Scott3142)
Think about what you need to do, you have a graph y=sinh(3x) and you need to find the area between two points, x=0 and x=ln(3). How would you usually calculate areas between points?
ah yes i forgot to mention ive already tried an integral between the interval 0 and ln3
it comes out as [(1/3)cosh3x] between 0 and ln3 then after a little working,
(1/3)(cosh(ln9) - 1) which i eventually got to be 32/27 which is apparently the wrong answer
4. It's not cosh(ln9) it'd be cosh(3ln3) = cosh(ln27)
5. (Original post by Scott3142)
It's not cosh(ln9) it'd be cosh(3ln3) = cosh(ln27)
ah thanks for pointing out

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Updated: December 9, 2010
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