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

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(Original post by

The answer is:

(sinxcoshx-cosxsinhx)/2

**mikesgt2**)The answer is:

(sinxcoshx-cosxsinhx)/2

Ben

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#22

(Original post by

i think hyperbolic functions are p5.

im only going up to p4, thank god :]

**kikzen**)i think hyperbolic functions are p5.

im only going up to p4, thank god :]

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#23

(Original post by

Thank the merciful lord, I've got it right!!!!!

Ben

**Ben.S.**)Thank the merciful lord, I've got it right!!!!!

Ben

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#24

(Original post by

Yeah, hyperbolics are in P5.

**Nylex**)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

Err OK will either of you care to share your working with the hyperbolically ignorant among us?

**ZJuwelH**)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

Well... if we tell you that:

d(sinhx)/dx = coshx

d(coshx)/dx = sinhx

You can work it out for yourself...

**mikesgt2**)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

Surely d(coshx)/dx = -sinhx

**Ralfskini**)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

**ZJuwelH**)

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

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,

**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,

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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

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

**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

Ben

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#37

(Original post by

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

**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

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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

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

**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

**DO**understand

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**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

Ben

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