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Arun Hallan
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What is the integration of sin2xsinx, and how do you do it?
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Omega
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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
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[q1]> What is the integration of sin2xsinx, and how do you do it?[/q1]
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Richard Hayden
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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
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[q1]> What is the integration of sin2xsinx, and how do you do it?[/q1]

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Richard Hayden
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[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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Stan Brown
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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."
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Darrell
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"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
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Dr A. N. Walker
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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]
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Stan Brown
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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)
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The Technical M
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"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]
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Robert Low
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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/
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Dr A. N. Walker
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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]
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Arun Hallan
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[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
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Richard Hayden
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[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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Arun Hallan
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"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]
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[q1]> Outgoing mail from Richard Hayden is certified Virus Free. Checked by AVG anti-virus system[/q1]
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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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Arun Hallan
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I've got a feeling there's gonna be a Q on newton's law of cooling...
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Richard Hayden
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[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.

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Arun Hallan
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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
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