# What's the integral of sin(x)sinh(x)?

This discussion is closed.
Thread starter 18 years ago
#21
(Original post by mikesgt2)

(sinxcoshx-cosxsinhx)/2
Thank the merciful lord, I've got it right!!!!!

Ben
0
18 years ago
#22
(Original post by kikzen)
i think hyperbolic functions are p5.

im only going up to p4, thank god :]
Yeah, hyperbolics are in P5. That's not even the worst bit of P5, Intrinsic Co-ordinates look worse.
0
18 years ago
#23
(Original post by Ben.S.)
Thank the merciful lord, I've got it right!!!!!

Ben
Err OK will either of you care to share your working with the hyperbolically ignorant among us?
0
18 years ago
#24
(Original post by Nylex)
Yeah, hyperbolics are in P5.
We learnt hyperbolics last week in further maths... I understand the definition etc...

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?
0
18 years ago
#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 .
0
Thread starter 18 years ago
#26
(Original post by ZJuwelH)
Err OK will either of you care to share your working with the hyperbolically ignorant among us?
You can have the working, if you like...

Ben
0
18 years ago
#27
Well... if we tell you that:

d(sinhx)/dx = coshx
d(coshx)/dx = sinhx

You can work it out for yourself...
0
18 years ago
#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...
Surely d(coshx)/dx = -sinhx
1
18 years ago
#29
Still not sure, I can differentiate but not integrate trig functions. Please show me the working, I'm willing to learn!
0
18 years ago
#30
(Original post by Ralfskini)
Surely d(coshx)/dx = -sinhx
Nope, it's sinh x, I just checked.
0
18 years ago
#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
0
18 years ago
#32
(Original post by ZJuwelH)
Err OK will either of you care to share your working with the hyperbolically ignorant among us?
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
18 years ago
#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,
Thanks Rich, but how does this translate into the answer that Ben and Mike got?
0
Thread starter 18 years ago
#34
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
0
18 years ago
#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
0
Thread starter 18 years ago
#36
(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
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
0
18 years ago
#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
<faints>
0
18 years ago
#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
0
18 years ago
#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
Now that I DO understand
0
Thread starter 18 years ago
#40
(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
lol - much better. Never mind - just a lot wasted typing time! Still - lots of good algebra practice...

Ben
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