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What's the integral of sin(x)sinh(x)?
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(Original post by mikesgt2)
The answer is:
(sinxcoshx-cosxsinhx)/2
The answer is:
(sinxcoshx-cosxsinhx)/2
Ben
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#22
(Original post by kikzen)
i think hyperbolic functions are p5.
im only going up to p4, thank god :]
i think hyperbolic functions are p5.
im only going up to p4, thank god :]
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#23
(Original post by Ben.S.)
Thank the merciful lord, I've got it right!!!!!
Ben
Thank the merciful lord, I've got it right!!!!!
Ben
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#24
(Original post by Nylex)
Yeah, hyperbolics are in P5.
Yeah, hyperbolics are in P5.
but, could you tell me what they are used for? The circular trig functions can be used for finding lengths and angles in a triangle, what are hyperbolic trig functions used for?
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#25
I dunno what hyperbolic functions are used for, I just taught myself a little bit of P5 before I left for uni (though I've forgotten it all now), soz 
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(Original post by ZJuwelH)
Err OK will either of you care to share your working with the hyperbolically ignorant among us?
Err OK will either of you care to share your working with the hyperbolically ignorant among us?
Ben
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#27
Well... if we tell you that:
d(sinhx)/dx = coshx
d(coshx)/dx = sinhx
You can work it out for yourself...
d(sinhx)/dx = coshx
d(coshx)/dx = sinhx
You can work it out for yourself...
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#28
(Original post by mikesgt2)
Well... if we tell you that:
d(sinhx)/dx = coshx
d(coshx)/dx = sinhx
You can work it out for yourself...
Well... if we tell you that:
d(sinhx)/dx = coshx
d(coshx)/dx = sinhx
You can work it out for yourself...
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#29
Still not sure, I can differentiate but not integrate trig functions. Please show me the working, I'm willing to learn!
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#30
(Original post by Ralfskini)
Surely d(coshx)/dx = -sinhx
Surely d(coshx)/dx = -sinhx
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#31
The definition of the hyperbolic functions is:
sinhx = ( e^x - e^-x )/2
coshx = ( e^x + e^-x )/2
You know that:
d(e^x)/dx = e^x
and d(e^-x)/dx = -e^-x
So:
d(coshx)/dx = ( d(e^x+e^-x)/dx )/2 = ( e^x - e^-x )/2 = sinhx
The hyperbolic trig functions behave slightly differently so while d(coshx)/dx = -sinhx, d(coshx)/dx = sinhx
sinhx = ( e^x - e^-x )/2
coshx = ( e^x + e^-x )/2
You know that:
d(e^x)/dx = e^x
and d(e^-x)/dx = -e^-x
So:
d(coshx)/dx = ( d(e^x+e^-x)/dx )/2 = ( e^x - e^-x )/2 = sinhx
The hyperbolic trig functions behave slightly differently so while d(coshx)/dx = -sinhx, d(coshx)/dx = sinhx
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#32
(Original post by ZJuwelH)
Err OK will either of you care to share your working with the hyperbolically ignorant among us?
Err OK will either of you care to share your working with the hyperbolically ignorant among us?
cosh(x) = cos(ix)
sinh(x) = -i*sin(ix)
Where i = sqrt(-1).
So we have d/dx(cosh(x)) = d/dx(cos(ix)) = -i*sin(ix) = -i*(i*sinh(x)) = sinh(x).
And also d/dx(sinh(x)) = d/dx(-i*sin(ix)) = -i*i*cos(ix) = cos(ix) = cosh(x).
Regards,
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#33
(Original post by rahaydenuk)
Hyperbolic functions are nothing special. Defined in terms of trigonometric functions, they are:
cosh(x) = cos(ix)
sinh(x) = -i*sin(ix)
Where i = sqrt(-1).
So we have d/dx(cosh(x)) = d/dx(cos(ix)) = -i*sin(ix) = -i*(i*sinh(x)) = sinh(x).
And also d/dx(sinh(x)) = d/dx(-i*sin(ix)) = -i*i*cos(ix) = cos(ix) = cosh(x).
Regards,
Hyperbolic functions are nothing special. Defined in terms of trigonometric functions, they are:
cosh(x) = cos(ix)
sinh(x) = -i*sin(ix)
Where i = sqrt(-1).
So we have d/dx(cosh(x)) = d/dx(cos(ix)) = -i*sin(ix) = -i*(i*sinh(x)) = sinh(x).
And also d/dx(sinh(x)) = d/dx(-i*sin(ix)) = -i*i*cos(ix) = cos(ix) = cosh(x).
Regards,
0
Int{sin(x)sinh(x)dx} =
= (1/4i)Int{[e^(ix) - e^(-ix)][e^(x) - e^(x)]dx}
Multiplying out the integrand, we have:
(1/4i)Int{e^[x(i + 1)] + e^[x(-i - 1) - e^[x(i - 1)] - e^[x(-i + 1)]dx}
Integrating and multiplying by the complex conjugates, top and bottom:
(1/8i){[-i + 1]e^(x(i + 1)) + [i - 1]e^(x(-i - 1)) - [-i - 1]e^(x(i - 1)) - [i + 1]e^(x(-i + 1))}
Multiplying out brackets and collecting real and imaginary parts:
(1/8i){e^(x)[e^(ix) - e^(-ix) - i[e^(ix) + e^(-ix)]] + e^(x)[e^(ix) - e^(-ix) + i[e^(ix) + e^(-ix)]]}
But, we have :
sin(x) = (1/2i)[e^(ix) - e^(-ix)]
and
cos(x) = (1/2)[e^(ix) + e^(-ix)]
So, we can write the big mess up there as:
(1/8i){e^(x)[2isin(x) - 2icos(x)] + e^(-x)[2isin(x) + 2icos(x)]}
Factorising out i and 2 gives:
(1/4){e^(x)[sin(x) - cos(x)] + e^(-x)[sin(x) + cos(x)]}
Rearranging we can get:
(1/4){sin(x)[e^(x) + e^(-x)] - cos(x)[e^(x) - e^(-x)]}
The hyperbolic functions are:
sinh(x) = (1/2)[e^(x) - e^(-x)]
and
cosh(x) = (1/2)[e^(x) + e^(-x)]
Substituting these identities in, we get:
(1/4)[2sin(x)cosh(x) - 2cos(x)sinh(x)]
So, finally, the integral is:
(1/2)[sin(x)cosh(x) - cos(x)sinh(x)] + C
Ben
= (1/4i)Int{[e^(ix) - e^(-ix)][e^(x) - e^(x)]dx}
Multiplying out the integrand, we have:
(1/4i)Int{e^[x(i + 1)] + e^[x(-i - 1) - e^[x(i - 1)] - e^[x(-i + 1)]dx}
Integrating and multiplying by the complex conjugates, top and bottom:
(1/8i){[-i + 1]e^(x(i + 1)) + [i - 1]e^(x(-i - 1)) - [-i - 1]e^(x(i - 1)) - [i + 1]e^(x(-i + 1))}
Multiplying out brackets and collecting real and imaginary parts:
(1/8i){e^(x)[e^(ix) - e^(-ix) - i[e^(ix) + e^(-ix)]] + e^(x)[e^(ix) - e^(-ix) + i[e^(ix) + e^(-ix)]]}
But, we have :
sin(x) = (1/2i)[e^(ix) - e^(-ix)]
and
cos(x) = (1/2)[e^(ix) + e^(-ix)]
So, we can write the big mess up there as:
(1/8i){e^(x)[2isin(x) - 2icos(x)] + e^(-x)[2isin(x) + 2icos(x)]}
Factorising out i and 2 gives:
(1/4){e^(x)[sin(x) - cos(x)] + e^(-x)[sin(x) + cos(x)]}
Rearranging we can get:
(1/4){sin(x)[e^(x) + e^(-x)] - cos(x)[e^(x) - e^(-x)]}
The hyperbolic functions are:
sinh(x) = (1/2)[e^(x) - e^(-x)]
and
cosh(x) = (1/2)[e^(x) + e^(-x)]
Substituting these identities in, we get:
(1/4)[2sin(x)cosh(x) - 2cos(x)sinh(x)]
So, finally, the integral is:
(1/2)[sin(x)cosh(x) - cos(x)sinh(x)] + C
Ben
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#35
Once again, wha-wha-wha-whaaaaaaaa??? Got confused after line 1, didn't bother reading the rest. But you're answer's right by the way 
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(Original post by ZJuwelH)
Once again, wha-wha-wha-whaaaaaaaa??? Got confused after line 1, didn't bother reading the rest. But you're answer's right by the way
Once again, wha-wha-wha-whaaaaaaaa??? Got confused after line 1, didn't bother reading the rest. But you're answer's right by the way
Ben
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#37
(Original post by Ben.S.)
Well, to be honest, I should have written down more than that - since I did quite a few steps at once. I couldn't be bothered to type more, though!
Ben
Well, to be honest, I should have written down more than that - since I did quite a few steps at once. I couldn't be bothered to type more, though!
Ben
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#38
u = sinx
u' = cosx
v' = sinhx
v = coshx
(int) sinx sinhx = sinx coshx - int cosx coshx
u = cosx
u'= -sinx
v' = coshx
v= sinhx
(int) sinx sinhx = sinx coshx - cosx sinhx - (int) sinx sinhx
2 (int) sinx sinhx = sinx coshx - cosx sinhx
=> (int) sinx sinhx = 1/2 (sinx coshx - cosx sinhx) +C
u' = cosx
v' = sinhx
v = coshx
(int) sinx sinhx = sinx coshx - int cosx coshx
u = cosx
u'= -sinx
v' = coshx
v= sinhx
(int) sinx sinhx = sinx coshx - cosx sinhx - (int) sinx sinhx
2 (int) sinx sinhx = sinx coshx - cosx sinhx
=> (int) sinx sinhx = 1/2 (sinx coshx - cosx sinhx) +C
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#39
(Original post by elpaw)
u = sinx
u' = cosx
v' = sinhx
v = coshx
(int) sinx sinhx = sinx coshx - int cosx coshx
u = cosx
u'= -sinx
v' = coshx
v= sinhx
(int) sinx sinhx = sinx coshx - cosx sinhx - (int) sinx sinhx
2 (int) sinx sinhx = sinx coshx - cosx sinhx
=> (int) sinx sinhx = 1/2 (sinx coshx - cosx sinhx) +C
u = sinx
u' = cosx
v' = sinhx
v = coshx
(int) sinx sinhx = sinx coshx - int cosx coshx
u = cosx
u'= -sinx
v' = coshx
v= sinhx
(int) sinx sinhx = sinx coshx - cosx sinhx - (int) sinx sinhx
2 (int) sinx sinhx = sinx coshx - cosx sinhx
=> (int) sinx sinhx = 1/2 (sinx coshx - cosx sinhx) +C
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(Original post by elpaw)
u = sinx
u' = cosx
v' = sinhx
v = coshx
(int) sinx sinhx = sinx coshx - int cosx coshx
u = cosx
u'= -sinx
v' = coshx
v= sinhx
(int) sinx sinhx = sinx coshx - cosx sinhx - (int) sinx sinhx
2 (int) sinx sinhx = sinx coshx - cosx sinhx
=> (int) sinx sinhx = 1/2 (sinx coshx - cosx sinhx) +C
u = sinx
u' = cosx
v' = sinhx
v = coshx
(int) sinx sinhx = sinx coshx - int cosx coshx
u = cosx
u'= -sinx
v' = coshx
v= sinhx
(int) sinx sinhx = sinx coshx - cosx sinhx - (int) sinx sinhx
2 (int) sinx sinhx = sinx coshx - cosx sinhx
=> (int) sinx sinhx = 1/2 (sinx coshx - cosx sinhx) +C
Ben
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