**INTEGRATION -- Urgent help please!** Watch

devesh254
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1) ∫ex sinhx dx

2) ∫sinh2 x dx between the limits 1 and 0

Thanks in advance!
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killerbee
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add one to the power and divide by the new power
sorry mate i never got p4 integration

try using some trig idntities
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devesh254
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hmmm that doesnt really help...

anyone else have any ideas?
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apd35
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1) by parts
2) use the substitution u = cosh x (du/dx=sinh x)
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meathead
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(Original post by devesh254)
1) ∫ex sinhx dx

2) ∫sinh2 x dx between the limits 1 and 0

Thanks in advance!
1) sinhx = ½(ex - e-x)
exsinhx = ½(e2x - 1)
∫½(e2x - 1)dx = ¼e2x - ½x + k

I don't think this one can be done by parts, because the sin/cos are hyperbolic you don't get any sign changes so you end up with the integrals cancelling ( I = excosh x - exsinh x + I )

2)sinh²x = ½(cosh 2x-1)

∫sinh²x dx = ½∫(cosh 2x-1)dx = ¼cosh 2x - ½x + k
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apd35
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(Original post by devesh254)
1) ∫ex sinhx dx
(dv/dx=ex u=sinh x
∫ex sinhx dx = ex coshx dx - ∫ex sinhx
2∫ex sinhx =ex coshx
∫ex sinhx = (1/2) ex coshx
I think that's right.
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devesh254
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(Original post by meathead)
1) sinhx = ½(ex - e-x)
exsinhx = ½(e2x - 1)
∫½(e2x - 1)dx = ¼e2x - ½x + k

2)sinh²x = ½(cosh 2x-1)

∫sinh²x dx = ½∫(cosh 2x-1)dx = ¼cosh 2x - ½x + k
cheers mate
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meathead
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(Original post by devesh254)
1) ∫ex sinhx dx

2) ∫sinh2 x dx between the limits 1 and 0

Thanks in advance!
I = ∫ex sinhx dx

Integration by parts:

I = ex coshx dx - ∫ex coshx dx

Now ∫ex coshx dx = ex cosh dx - ∫ex sinhx dx

∴ I = ex coshx dx - ∫ex coshx dx
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devesh254
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(Original post by meathead)

∫sinh²x dx = ½∫(cosh 2x-1)dx
are there any more trig identities like the above which are useful with integration? is there any other form for cosh²x?

thanks!
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Dekota
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(Original post by devesh254)
are there any more trig identities like the above which are useful with integration? is there any other form for cosh²x?

thanks!
http://www.answers.com/trig%20identities
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devesh254
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(Original post by Dekota)
http://www.answers.com/trig%20identities
so u can use the double angle formulas, i.e. sin2x and cox2x, for sinh2x and cosh2x?

so sinh2x = 2sinhxcoshx
cosh2x = cosh²x - sinh²x = 2cosh²x - 1 = 1 - 2sinh²x

is the above correct?

thanks!
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apd35
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cosh2x = cosh²x + sinh²x = 2cosh²x - 1 = 1 + 2sinh²x
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yusufu
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(Original post by devesh254)
so u can use the double angle formulas, i.e. sin2x and cox2x, for sinh2x and cosh2x?

so sinh2x = 2sinhxcoshx
cosh2x = cosh²x - sinh²x = 2cosh²x - 1 = 1 - 2sinh²x

is the above correct?

thanks!
youre using trigonometric identities rahter than hyperbolic identities.
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