# Integration of sin[2x]sin[x]

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#2

integration of sin2xsinx

= 2sinXcosXsinX 2sin^2 X cosX

let u = sin^2 X v' = cosX

u' = sin2x v= sinX

integration of sinXsin2X = sinX sin^2 X - integrate of sinX sin2X

2 * integration of sinXsin2x = sin^3 X

integration of sinXsin2X = 1/2 (sin^3 X)

understand??

Omega

"Arun Hallan" <[email protected]> wrote in message

news:[email protected]...

[q1]> What is the integration of sin2xsinx, and how do you do it?[/q1]

= 2sinXcosXsinX 2sin^2 X cosX

let u = sin^2 X v' = cosX

u' = sin2x v= sinX

integration of sinXsin2X = sinX sin^2 X - integrate of sinX sin2X

2 * integration of sinXsin2x = sin^3 X

integration of sinXsin2X = 1/2 (sin^3 X)

understand??

Omega

"Arun Hallan" <[email protected]> wrote in message

news:[email protected]...

[q1]> What is the integration of sin2xsinx, and how do you do it?[/q1]

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#3

Hi,

This can be done by a bit of manipulation and then can be integrated by sight. There's not really

any need for integration by substitution or parts.

Recall that, Sin(2a) = 2Sin(a)Cos(a)

Int(Sin(2x)Sin(x)) dx = Int(2(Sin(x)^2)Cos(x)) dx

You should then be able to see easily that the basic form of the result is going to be:

nSin(x)^3

And a bit more thinking will lead you to see that n should be 2/3, giving:

Int(Sin(2x)Sin(x)) dx = (2/3)Sin(x)^3

This could probably have been done by substitution or by parts, but I find it much more fun to do it

'by sight'.

Regards,

Richard Hayden.

"Arun Hallan" <[email protected]> wrote in message

news:[email protected]...

[q1]> What is the integration of sin2xsinx, and how do you do it?[/q1]

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This can be done by a bit of manipulation and then can be integrated by sight. There's not really

any need for integration by substitution or parts.

Recall that, Sin(2a) = 2Sin(a)Cos(a)

Int(Sin(2x)Sin(x)) dx = Int(2(Sin(x)^2)Cos(x)) dx

You should then be able to see easily that the basic form of the result is going to be:

nSin(x)^3

And a bit more thinking will lead you to see that n should be 2/3, giving:

Int(Sin(2x)Sin(x)) dx = (2/3)Sin(x)^3

This could probably have been done by substitution or by parts, but I find it much more fun to do it

'by sight'.

Regards,

Richard Hayden.

"Arun Hallan" <[email protected]> wrote in message

news:[email protected]...

[q1]> What is the integration of sin2xsinx, and how do you do it?[/q1]

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#4

[q1]> integration of sin2xsinx[/q1]

[q1]>[/q1]

[q1]> = 2sinXcosXsinX 2sin^2 X cosX[/q1]

[q1]>[/q1]

[q1]> let u = sin^2 X v' = cosX[/q1]

[q1]> u' = sin2x v= sinX[/q1]

[q1]>[/q1]

[q1]> integration of sinXsin2X = sinX sin^2 X - integrate of sinX sin2X[/q1]

[q1]>[/q1]

[q1]> 2 * integration of sinXsin2x = sin^3 X[/q1]

[q1]>[/q1]

[q1]> integration of sinXsin2X = 1/2 (sin^3 X)[/q1]

[q1]>[/q1]

[q1]> understand??[/q1]

[q1]>[/q1]

[q1]> Omega[/q1]

I'm afraid that's incorrect. The correct answer is (2/3)Sin(x)^3.

See my other post in this thread for my method. Integration by parts isn't really needed anyway!

Regards,

Richard Hayden.

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[q1]>[/q1]

[q1]> = 2sinXcosXsinX 2sin^2 X cosX[/q1]

[q1]>[/q1]

[q1]> let u = sin^2 X v' = cosX[/q1]

[q1]> u' = sin2x v= sinX[/q1]

[q1]>[/q1]

[q1]> integration of sinXsin2X = sinX sin^2 X - integrate of sinX sin2X[/q1]

[q1]>[/q1]

[q1]> 2 * integration of sinXsin2x = sin^3 X[/q1]

[q1]>[/q1]

[q1]> integration of sinXsin2X = 1/2 (sin^3 X)[/q1]

[q1]>[/q1]

[q1]> understand??[/q1]

[q1]>[/q1]

[q1]> Omega[/q1]

I'm afraid that's incorrect. The correct answer is (2/3)Sin(x)^3.

See my other post in this thread for my method. Integration by parts isn't really needed anyway!

Regards,

Richard Hayden.

---

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#5

Arun Hallan <[email protected]> wrote in uk.education.maths:

[q1]>What is the integration of sin2xsinx, and how do you do it?[/q1]

Apply the double-angle formula to expand sin(2x). Then make a substitution u = sin x. When you

figure what du is, it should be obvious how to proceed.

--

Stan Brown, Oak Road Systems, Cortland County, New York, USA http://oakroadsystems.com/ "What in

heaven's name brought you to Casablanca?" "My health. I came to Casablanca for the waters." "The

waters? What waters? We're in the desert." "I was misinformed."

[q1]>What is the integration of sin2xsinx, and how do you do it?[/q1]

Apply the double-angle formula to expand sin(2x). Then make a substitution u = sin x. When you

figure what du is, it should be obvious how to proceed.

--

Stan Brown, Oak Road Systems, Cortland County, New York, USA http://oakroadsystems.com/ "What in

heaven's name brought you to Casablanca?" "My health. I came to Casablanca for the waters." "The

waters? What waters? We're in the desert." "I was misinformed."

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#6

"Omega" <[email protected]> wrote in message

news:[email protected]...

[q1]>[/q1]

[q1]>[/q1]

[q1]> integration of sin2xsinx[/q1]

[q1]>[/q1]

[q1]> = 2sinXcosXsinX 2sin^2 X cosX[/q1]

[q1]>[/q1]

[q1]> let u = sin^2 X v' = cosX[/q1]

[q1]> u' = sin2x v= sinX[/q1]

Parts? This is just a power rule...

int[u^n du] = u^(n+1)/(n+1) + C, n <> -1

u = sin(x) du = cos(x) dx

So, 2int[sin^2(x) cos(x) dx]

= 2sin^3(x)/ 3 + C

Your attempt at parts has the incorrect u'. Keeping everything the the same except using the correct

du, we have:

u = sin^2(x) dv = cos(x) dx du = 2sin(x)cos(x) dx v = sin(x)

int[u dv] = uv - int[v du]

int[sin^2(x)cos(x) dx] = sin^3(x) - int[2sin^2(x)cos(x) dx]

= sin^3(x) - 2 int[sin^2(x)cos(x) dx]

...just pull out the 2

Hopefully you can easily see why I pulled out the 2. This can now be evaluated without performing

any integration. Just algebraically solve this equation for int[sin^2(x)cos(x)dx]. Of course the

result needs to be multipied by 2, since the integrand is twice your choice for u*dv.

If you are going to use parts, in general it's usually best to let dv be the most complicated

portion that fits a basic integration rule, and let u be the portion whose derivative is simpler

than u. Both conditions can be met by letting:

dv = sin^2(x)cos(x) dx u = 2

v = sin^3(x) / 3 du = 0 dx

int[ u dv] = uv - int[v du]

int[2sin^2(x)cos(x) dx] = 2sin^3(x)/3 - int[0 dx]

= 2sin^3(x)/3 + C

Of course, this example is kind of moot since the integrand itself "fits" a basic integration rule

to begin with, thus no need for parts.

--

Darrell

news:[email protected]...

[q1]>[/q1]

[q1]>[/q1]

[q1]> integration of sin2xsinx[/q1]

[q1]>[/q1]

[q1]> = 2sinXcosXsinX 2sin^2 X cosX[/q1]

[q1]>[/q1]

[q1]> let u = sin^2 X v' = cosX[/q1]

[q1]> u' = sin2x v= sinX[/q1]

Parts? This is just a power rule...

int[u^n du] = u^(n+1)/(n+1) + C, n <> -1

u = sin(x) du = cos(x) dx

So, 2int[sin^2(x) cos(x) dx]

= 2sin^3(x)/ 3 + C

Your attempt at parts has the incorrect u'. Keeping everything the the same except using the correct

du, we have:

u = sin^2(x) dv = cos(x) dx du = 2sin(x)cos(x) dx v = sin(x)

int[u dv] = uv - int[v du]

int[sin^2(x)cos(x) dx] = sin^3(x) - int[2sin^2(x)cos(x) dx]

= sin^3(x) - 2 int[sin^2(x)cos(x) dx]

...just pull out the 2

Hopefully you can easily see why I pulled out the 2. This can now be evaluated without performing

any integration. Just algebraically solve this equation for int[sin^2(x)cos(x)dx]. Of course the

result needs to be multipied by 2, since the integrand is twice your choice for u*dv.

If you are going to use parts, in general it's usually best to let dv be the most complicated

portion that fits a basic integration rule, and let u be the portion whose derivative is simpler

than u. Both conditions can be met by letting:

dv = sin^2(x)cos(x) dx u = 2

v = sin^3(x) / 3 du = 0 dx

int[ u dv] = uv - int[v du]

int[2sin^2(x)cos(x) dx] = 2sin^3(x)/3 - int[0 dx]

= 2sin^3(x)/3 + C

Of course, this example is kind of moot since the integrand itself "fits" a basic integration rule

to begin with, thus no need for parts.

--

Darrell

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#7

In article <[email protected] .odyssey.net>, Stan Brown <[email protected]> wrote:

[q1]>Arun Hallan <[email protected]> wrote in uk.education.maths:[/q1]

[q2]>>What is the integration of sin2xsinx, and how do you do it?[/q2]

[q1]>Apply the double-angle formula to expand sin(2x). Then make a substitution u = sin x. [...][/q1]

Alternatively, use the trig addition formulas:

cos x - cos 3x == 2 sin .5(3x+x) sin .5(3x-x) == 2 sin 2x sin x,

and the integral of the LHS is easy. Or, for those who know Euler's formulas, use

2i sin x == exp ix - exp -ix, 2i sin 2x == exp 2ix - exp -2ix,

multiply out and integrate the exponentials. Even easier, use Maple:

> int (sin(2*x)*sin(x), x);

1/2 sin(x) - 1/6 sin(3 x)

[Aside: 40-odd years ago, Atlas Autocode used to recognise "2a", "a.b" and "f(a)f(b)" as implicit

multiplications; shame that computer languages are, on the whole, less maths-friendly today than

they were then ....]

--

Andy Walker, School of MathSci., Univ. of Nott'm, UK. [email protected]

[q1]>Arun Hallan <[email protected]> wrote in uk.education.maths:[/q1]

[q2]>>What is the integration of sin2xsinx, and how do you do it?[/q2]

[q1]>Apply the double-angle formula to expand sin(2x). Then make a substitution u = sin x. [...][/q1]

Alternatively, use the trig addition formulas:

cos x - cos 3x == 2 sin .5(3x+x) sin .5(3x-x) == 2 sin 2x sin x,

and the integral of the LHS is easy. Or, for those who know Euler's formulas, use

2i sin x == exp ix - exp -ix, 2i sin 2x == exp 2ix - exp -2ix,

multiply out and integrate the exponentials. Even easier, use Maple:

> int (sin(2*x)*sin(x), x);

1/2 sin(x) - 1/6 sin(3 x)

[Aside: 40-odd years ago, Atlas Autocode used to recognise "2a", "a.b" and "f(a)f(b)" as implicit

multiplications; shame that computer languages are, on the whole, less maths-friendly today than

they were then ....]

--

Andy Walker, School of MathSci., Univ. of Nott'm, UK. [email protected]

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#8

Dr A. N. Walker <[email protected]> wrote in uk.education.maths:

[q1]>[Aside: 40-odd years ago, Atlas Autocode used to recognise "2a", "a.b" and "f(a)f(b)" as implicit[/q1]

[q1]>multiplications; shame that computer languages are, on the whole, less maths-friendly today than[/q1]

[q1]>they were then ....][/q1]

So did APL.

But the restriction (wwhich also applies to your example) was that a variable had to be a single

letter. That is hardly maths-friendly.

Consider writing a program to solve triangles by the law of sines or law of cosines (as I recently

did for the TI-83). Is "a" side a or angle A? It's hard to keep track when you can have only single-

letter variables.

--

Stan Brown, Oak Road Systems, Cortland County, New York, USA http://oakroadsystems.com/ "My theory

was a perfectly good one. The facts were misleading." -- /The Lady Vanishes/ (1938)

[q1]>[Aside: 40-odd years ago, Atlas Autocode used to recognise "2a", "a.b" and "f(a)f(b)" as implicit[/q1]

[q1]>multiplications; shame that computer languages are, on the whole, less maths-friendly today than[/q1]

[q1]>they were then ....][/q1]

So did APL.

But the restriction (wwhich also applies to your example) was that a variable had to be a single

letter. That is hardly maths-friendly.

Consider writing a program to solve triangles by the law of sines or law of cosines (as I recently

did for the TI-83). Is "a" side a or angle A? It's hard to keep track when you can have only single-

letter variables.

--

Stan Brown, Oak Road Systems, Cortland County, New York, USA http://oakroadsystems.com/ "My theory

was a perfectly good one. The facts were misleading." -- /The Lady Vanishes/ (1938)

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#9

"Dr A. N. Walker" wrote:

[q1]> In article <[email protected] .odyssey.net>, Stan Brown <[email protected]> wrote:[/q1]

[q2]> >Arun Hallan <[email protected]> wrote in uk.education.maths:[/q2]

[q2]> >>What is the integration of sin2xsinx, and how do you do it?[/q2]

[q2]> >Apply the double-angle formula to expand sin(2x). Then make a substitution u = sin x. [...][/q2]

[q1]>[/q1]

[q1]> Alternatively, use the trig addition formulas:[/q1]

[q1]>[/q1]

[q1]> cos x - cos 3x == 2 sin .5(3x+x) sin .5(3x-x) == 2 sin 2x sin x,[/q1]

[q1]>[/q1]

[q1]> and the integral of the LHS is easy. Or, for those who know Euler's formulas, use[/q1]

[q1]>[/q1]

[q1]> 2i sin x == exp ix - exp -ix, 2i sin 2x == exp 2ix - exp -2ix,[/q1]

[q1]>[/q1]

[q1]> multiply out and integrate the exponentials.[/q1]

Thats the real smart technique. Sometimes there is more than one way to solve a problem. Out of

interest do bonus marks exist for A level maths ?

[q1]> Even easier, use Maple:[/q1]

[q1]>[/q1]

[q1]> > int (sin(2*x)*sin(x), x);[/q1]

[q1]> 1/2 sin(x) - 1/6 sin(3 x)[/q1]

There are times when I wonder whether its really worth bothering including all these different

techniques to integrate products and quotients at A Level anymore now that software like Maple

exists. As an engineer I think that the time would be better spent teaching numerical methods or

mathematical modelling instead. Many functions I deal with can't be integrated.

[q1]>[/q1]

[q1]>[/q1]

[q1]> [Aside: 40-odd years ago, Atlas Autocode used to recognise "2a", "a.b" and "f(a)f(b)" as implicit[/q1]

[q1]> multiplications; shame that computer languages are, on the whole, less maths-friendly today than[/q1]

[q1]> they were then ....][/q1]

[q1]>[/q1]

[q1]> --[/q1]

[q1]> Andy Walker, School of MathSci., Univ. of Nott'm, UK. [email protected][/q1]

[q1]> In article <[email protected] .odyssey.net>, Stan Brown <[email protected]> wrote:[/q1]

[q2]> >Arun Hallan <[email protected]> wrote in uk.education.maths:[/q2]

[q2]> >>What is the integration of sin2xsinx, and how do you do it?[/q2]

[q2]> >Apply the double-angle formula to expand sin(2x). Then make a substitution u = sin x. [...][/q2]

[q1]>[/q1]

[q1]> Alternatively, use the trig addition formulas:[/q1]

[q1]>[/q1]

[q1]> cos x - cos 3x == 2 sin .5(3x+x) sin .5(3x-x) == 2 sin 2x sin x,[/q1]

[q1]>[/q1]

[q1]> and the integral of the LHS is easy. Or, for those who know Euler's formulas, use[/q1]

[q1]>[/q1]

[q1]> 2i sin x == exp ix - exp -ix, 2i sin 2x == exp 2ix - exp -2ix,[/q1]

[q1]>[/q1]

[q1]> multiply out and integrate the exponentials.[/q1]

Thats the real smart technique. Sometimes there is more than one way to solve a problem. Out of

interest do bonus marks exist for A level maths ?

[q1]> Even easier, use Maple:[/q1]

[q1]>[/q1]

[q1]> > int (sin(2*x)*sin(x), x);[/q1]

[q1]> 1/2 sin(x) - 1/6 sin(3 x)[/q1]

There are times when I wonder whether its really worth bothering including all these different

techniques to integrate products and quotients at A Level anymore now that software like Maple

exists. As an engineer I think that the time would be better spent teaching numerical methods or

mathematical modelling instead. Many functions I deal with can't be integrated.

[q1]>[/q1]

[q1]>[/q1]

[q1]> [Aside: 40-odd years ago, Atlas Autocode used to recognise "2a", "a.b" and "f(a)f(b)" as implicit[/q1]

[q1]> multiplications; shame that computer languages are, on the whole, less maths-friendly today than[/q1]

[q1]> they were then ....][/q1]

[q1]>[/q1]

[q1]> --[/q1]

[q1]> Andy Walker, School of MathSci., Univ. of Nott'm, UK. [email protected][/q1]

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#10

Stan Brown <[email protected]> wrote:

[q1]>Consider writing a program to solve triangles by the law of sines or law of cosines (as I recently[/q1]

[q1]>did for the TI-83). Is "a" side a or angle A? It's hard to keep track when you can have only[/q1]

[q1]>single- letter variables.[/q1]

Not if it's case-sensitive :-)

--

Rob. http://www.mis.coventry.ac.uk/~mtx014/

[q1]>Consider writing a program to solve triangles by the law of sines or law of cosines (as I recently[/q1]

[q1]>did for the TI-83). Is "a" side a or angle A? It's hard to keep track when you can have only[/q1]

[q1]>single- letter variables.[/q1]

Not if it's case-sensitive :-)

--

Rob. http://www.mis.coventry.ac.uk/~mtx014/

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#11

In article <[email protected] .odyssey.net>, Stan Brown <[email protected]> wrote:

[q1]>Dr A. N. Walker <[email protected]> wrote in uk.education.maths:[/q1]

[q2]>>[Aside: 40-odd years ago, Atlas Autocode used to recognise "2a", "a.b" and "f(a)f(b)" as implicit[/q2]

[q2]>>multiplications; shame that computer languages are, on the whole, less maths-friendly today than[/q2]

[q2]>>they were then ....][/q2]

[q1]>So did APL. But the restriction (wwhich also applies to your example) was that a variable had to be[/q1]

[q1]>a single letter. That is hardly maths-friendly.[/q1]

Hmm! (a) Atlas Autocode did *not* have this restriction: "momentum = mass.velocity" or

"sin(2x)sin(x)" or "2fred123.jim", for example. Of course, "2.3" was two-and-a-bit

rather than six!

(b) OTOH, maths, exc for standard functions, by and large, *does* have this restriction: "Let fred

be the number of sheep" just doesn't sound right!

[q1]>Consider writing a program to solve triangles by the law of sines or law of cosines (as I recently[/q1]

[q1]>did for the TI-83). Is "a" side a or angle A? It's hard to keep track when you can have only[/q1]

[q1]>single- letter variables.[/q1]

Also in Atlas Autocode, "A" and "a" were different variables, unless you set "upper case

delimiters" in which case "A" was an unknown "reserved word" and therefore illegal.

--

Andy Walker, School of MathSci., Univ. of Nott'm, UK. [email protected]

[q1]>Dr A. N. Walker <[email protected]> wrote in uk.education.maths:[/q1]

[q2]>>[Aside: 40-odd years ago, Atlas Autocode used to recognise "2a", "a.b" and "f(a)f(b)" as implicit[/q2]

[q2]>>multiplications; shame that computer languages are, on the whole, less maths-friendly today than[/q2]

[q2]>>they were then ....][/q2]

[q1]>So did APL. But the restriction (wwhich also applies to your example) was that a variable had to be[/q1]

[q1]>a single letter. That is hardly maths-friendly.[/q1]

Hmm! (a) Atlas Autocode did *not* have this restriction: "momentum = mass.velocity" or

"sin(2x)sin(x)" or "2fred123.jim", for example. Of course, "2.3" was two-and-a-bit

rather than six!

(b) OTOH, maths, exc for standard functions, by and large, *does* have this restriction: "Let fred

be the number of sheep" just doesn't sound right!

[q1]>Consider writing a program to solve triangles by the law of sines or law of cosines (as I recently[/q1]

[q1]>did for the TI-83). Is "a" side a or angle A? It's hard to keep track when you can have only[/q1]

[q1]>single- letter variables.[/q1]

Also in Atlas Autocode, "A" and "a" were different variables, unless you set "upper case

delimiters" in which case "A" was an unknown "reserved word" and therefore illegal.

--

Andy Walker, School of MathSci., Univ. of Nott'm, UK. [email protected]

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#12

[email protected] (Arun Hallan) wrote in message

news:<b29468d2.0205110616.1d5f9c [email protected]>...

[q1]> What is the integration of sin2xsinx, and how do you do it?[/q1]

Thanx all, figured it out anyway...

i did cos(A+B) - cos(A-B) where A=2x and B=x

cos(A+B) = cosAcosB - sinAsinB cos(A-B) = cosAcosB + sinAsinB

cos(A+B) - cos(A-B) = -2sinAsinB cos 3x - cos x = -2sinAsinB

and it should come out as (sin[2x]sin[x]) = {cos[3x] - cos[x]} divide by -2

... much easier

If any1 was wondering, the Q is off Q8 on the edexcel P3 exam from June 2001 ... got the real one

on monday

anyway thanx

news:<b29468d2.0205110616.1d5f9c [email protected]>...

[q1]> What is the integration of sin2xsinx, and how do you do it?[/q1]

Thanx all, figured it out anyway...

i did cos(A+B) - cos(A-B) where A=2x and B=x

cos(A+B) = cosAcosB - sinAsinB cos(A-B) = cosAcosB + sinAsinB

cos(A+B) - cos(A-B) = -2sinAsinB cos 3x - cos x = -2sinAsinB

and it should come out as (sin[2x]sin[x]) = {cos[3x] - cos[x]} divide by -2

... much easier

If any1 was wondering, the Q is off Q8 on the edexcel P3 exam from June 2001 ... got the real one

on monday

anyway thanx

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#13

[q1]> If any1 was wondering, the Q is off Q8 on the edexcel P3 exam from June 2001 ... got the real one[/q1]

[q1]> on monday [/q1]

Thought I recognised it from somewhere.....

I've got that exam on Monday as well, it should be fun...

You should try the last question on the January 2002 Edexcel P3 exam, that's a bit of light

entertainment.

Good Luck!!

Regards,

Richard Hayden.

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[q1]> on monday [/q1]

Thought I recognised it from somewhere.....

I've got that exam on Monday as well, it should be fun...

You should try the last question on the January 2002 Edexcel P3 exam, that's a bit of light

entertainment.

Good Luck!!

Regards,

Richard Hayden.

---

Outgoing mail from Richard Hayden is certified Virus Free. Checked by AVG anti-virus system

(http://www.grisoft.com). Version: 6.0.317 / Virus Database: 176 - Release Date: 21/01/2002

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#14

"Richard Hayden" <[email protected]> wrote in message

news:<[email protected] l.co.uk>...

[q2]> > If any1 was wondering, the Q is off Q8 on the edexcel P3 exam from June 2001 ... got the real[/q2]

[q2]> > one on monday [/q2]

[q1]>[/q1]

[q1]> Thought I recognised it from somewhere.....[/q1]

[q1]>[/q1]

[q1]> I've got that exam on Monday as well, it should be fun...[/q1]

[q1]>[/q1]

[q1]> You should try the last question on the January 2002 Edexcel P3 exam, that's a bit of light[/q1]

[q1]> entertainment.[/q1]

[q1]>[/q1]

[q1]> Good Luck!![/q1]

[q1]>[/q1]

[q1]> Regards,[/q1]

[q1]>[/q1]

[q1]> Richard Hayden.[/q1]

[q1]>[/q1]

[q1]>[/q1]

[q1]> ---[/q1]

[q1]> Outgoing mail from Richard Hayden is certified Virus Free. Checked by AVG anti-virus system[/q1]

[q1]> (http://www.grisoft.com). Version: 6.0.317 / Virus Database: 176 - Release Date: 21/01/2002[/q1]

Yeh doing that paper right now actually, every1 ses the last Q is solid ... really hope the main

integration one on monday decides to be a bit easier

news:<[email protected] l.co.uk>...

[q2]> > If any1 was wondering, the Q is off Q8 on the edexcel P3 exam from June 2001 ... got the real[/q2]

[q2]> > one on monday [/q2]

[q1]>[/q1]

[q1]> Thought I recognised it from somewhere.....[/q1]

[q1]>[/q1]

[q1]> I've got that exam on Monday as well, it should be fun...[/q1]

[q1]>[/q1]

[q1]> You should try the last question on the January 2002 Edexcel P3 exam, that's a bit of light[/q1]

[q1]> entertainment.[/q1]

[q1]>[/q1]

[q1]> Good Luck!![/q1]

[q1]>[/q1]

[q1]> Regards,[/q1]

[q1]>[/q1]

[q1]> Richard Hayden.[/q1]

[q1]>[/q1]

[q1]>[/q1]

[q1]> ---[/q1]

[q1]> Outgoing mail from Richard Hayden is certified Virus Free. Checked by AVG anti-virus system[/q1]

[q1]> (http://www.grisoft.com). Version: 6.0.317 / Virus Database: 176 - Release Date: 21/01/2002[/q1]

Yeh doing that paper right now actually, every1 ses the last Q is solid ... really hope the main

integration one on monday decides to be a bit easier

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#16

[q1]> Yeh doing that paper right now actually, every1 ses the last Q is solid ... really hope the main[/q1]

[q1]> integration one on monday decides to be a bit easier[/q1]

I managed to do it with a bit of thought, but no one else in my class could. It just seems hard as

it is using algebra rather than numbers. You also have to be OK with parametric integration.

[q1]> I've got a feeling there's gonna be a Q on newton's law of cooling...[/q1]

Quite possibly! I don't think there's been one of those for a while.

Good Luck!!

Richard Hayden.

---

Outgoing mail from Richard Hayden is certified Virus Free. Checked by AVG anti-virus system

(http://www.grisoft.com). Version: 6.0.317 / Virus Database: 176 - Release Date: 21/01/2002

[q1]> integration one on monday decides to be a bit easier[/q1]

I managed to do it with a bit of thought, but no one else in my class could. It just seems hard as

it is using algebra rather than numbers. You also have to be OK with parametric integration.

[q1]> I've got a feeling there's gonna be a Q on newton's law of cooling...[/q1]

Quite possibly! I don't think there's been one of those for a while.

Good Luck!!

Richard Hayden.

---

Outgoing mail from Richard Hayden is certified Virus Free. Checked by AVG anti-virus system

(http://www.grisoft.com). Version: 6.0.317 / Virus Database: 176 - Release Date: 21/01/2002

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#17

Yeh ive cracked parametric integration, it's just using the substitution method... know how to find

out the limits aswel, and other things that are simular in the two past Q8's

Good Luck Richard

out the limits aswel, and other things that are simular in the two past Q8's

Good Luck Richard

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