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INTEGRATION -- Urgent help please!

devesh254

1) ∫e^x sinhx dx

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

Thanks in advance!

Reply 1

killerbee

add one to the power and divide by the new power
sorry mate i never got p4 integration

try using some trig idntities

Reply 2

devesh254

hmmm that doesnt really help...

anyone else have any ideas?

Reply 3

apd35

1) by parts
2) use the substitution u = cosh x (du/dx=sinh x)

Reply 4

meathead

devesh254

1) ∫e^x sinhx dx

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

Thanks in advance!

1) sinhx = ½(e^x - e^-x)
e^xsinhx = ½(e^2x - 1)
∫½(e^2x - 1)dx = ¼e^2x - 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 = e^xcosh x - e^xsinh x + I )

2)sinh²x = ½(cosh 2x-1)

∫sinh²x dx = ½∫(cosh 2x-1)dx = ¼cosh 2x - ½x + k

Reply 5

apd35

devesh254

1) ∫e^x sinhx dx

(dv/dx=e^x u=sinh x
∫e^x sinhx dx = e^x coshx dx - ∫e^x sinhx
2∫e^x sinhx =e^x coshx
∫e^x sinhx = (1/2) e^x coshx
I think that's right.

Reply 6

devesh254

meathead

1) sinhx = ½(e^x - e^-x)
e^xsinhx = ½(e^2x - 1)
∫½(e^2x - 1)dx = ¼e^2x - ½x + k

2)sinh²x = ½(cosh 2x-1)

∫sinh²x dx = ½∫(cosh 2x-1)dx = ¼cosh 2x - ½x + k

cheers mate

Reply 7

meathead

devesh254

1) ∫e^x sinhx dx

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

Thanks in advance!

I = ∫e^x sinhx dx

Integration by parts:

I = e^x coshx dx - ∫e^x coshx dx

Now ∫e^x coshx dx = e^x cosh dx - ∫e^x sinhx dx

∴ I = e^x coshx dx - ∫e^x coshx dx

Reply 8

devesh254

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!

Reply 9

Dekota

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

Reply 10

devesh254

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!

Reply 11

apd35

cosh2x = cosh²x + sinh²x = 2cosh²x - 1 = 1 + 2sinh²x

Reply 12

yusufu

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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