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    (Original post by RDKGames)
    Because the constants are in different places I suppose...

    \frac{d}{dx}[f(x)]^n = nf'(x)[f(x)]^{n-1} is the form you know (hopefully). So if n \mapsto (n+1) we have:

    \frac{d}{dx}[f(x)]^{n+1}=(n+1)f'(x)[f(x)]^{n} so dividing both sides by (n+1) gives \frac{1}{n+1}\frac{d}{dx}[f(x)]^{n+1}=f'(x)[f(x)]^{n}

    Integrating both sides gives \frac{1}{n+1}[f(x)]^{n+1} = \int f'(x)[f(x)]^{n} .dx
    Ok so I may have to read that a few more times as I'm having trouble understanding as I only know the chain rule as  \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}
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    (Original post by Philip-flop)
    Ok so I may have to read that a few more times as I'm having trouble understanding as I only know the chain rule as  \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}
    Well I wasn't deriving it using the chain rule here, I derived it from the form you know...
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    (Original post by RDKGames)
    Well I wasn't deriving it using the chain rule here, I derived it from the form you know...
    Ok so I can see how \frac{d}{dx}[f(x)]^n = nf'(x)[f(x)]^{n-1} is the Chain Rule now aka  \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}

    And I can follow how the inspection method of integrating is found aka \frac{1}{n+1}[f(x)]^{n+1} = \int f'(x)[f(x)]^{n} .dx
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    So I just came across a random question online...
     \int 2sin(3x)sin(2x) dx

    Under the Edexcel spec, are we expected to know that from Double Angle Formulae that we can derive the fact that?...
     cos(a-b) - cos(a+b) \equiv 2sin(a)sin(b)
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    (Original post by Philip-flop)
    So I just came across a random question online...
     \int 2sin(3x)sin(2x) dx

    Under the Edexcel spec, are we expected to know that from Double Angle Formulae that we can derive the fact that?...
     cos(a-b) - cos(a+b) \equiv 2sin(a)sin(b)
    Not really but I guess you can use it.

    Otherwise, write sin(3x)=sin(2x+x) and expand
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    (Original post by RDKGames)
    Not really but I guess you can use it.

    Otherwise, write sin(3x)=sin(2x+x) and expand
    Just looked at the specification: is there really no mention of sum-to-product and product-to-sum formulae? Have to say, I'm shocked and appalled...
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    (Original post by RDKGames)
    Not really but I guess you can use it.

    Otherwise, write sin(3x)=sin(2x+x) and expand
    Awesome. Thank you!
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    (Original post by DFranklin)
    Just looked at the specification: is there really no mention of sum-to-product and product-to-sum formulae? Have to say, I'm shocked and appalled...
    Didn't do Edexcel but it was never mentioned in any pure modules, even FP, though those formulae are just there in the formula booklet
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    (Original post by RDKGames)
    Not really but I guess you can use it.

    Otherwise, write sin(3x)=sin(2x+x) and expand
    So I still struggled to solve  \int 2sin(3x)sin(2x)

    I tried to expand...
     sin(3x) = sin(2x+x) = sin(2x)cos(x) + cos(2x) sin(x)

    But then didn't really know how to use it from there as I don't really know what I'm doing
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    (Original post by Philip-flop)
    So I still struggled to solve  \int 2sin(3x)sin(2x)

    I tried to expand...
     sin(3x) = sin(2x+x) = sin(2x)cos(x) + cos(2x) sin(x)

    But then didn't really know how to use it from there as I don't really know what I'm doing
    The best way to do this is to use the factor formula in the formula booklet:

    \displaystyle \cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}

    There is an exercise in the textbook that uses these "factor formulae".

    I'm guessing this wasn't an exam question? Even though they're in the formula booklet, they're not really seen in Edexcel exams. Just be aware that when you have a product of sin or cos like this then you may be able to use the factor formulae.
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    (Original post by notnek)
    The best way to do this is to use the factor formula in the formula booklet:

    \displaystyle \cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}

    There is an exercise in the textbook that uses these "factor formulae".

    I'm guessing this wasn't an exam question? Even though they're in the formula booklet, they're not really seen in Edexcel exams. Just be aware that when you have a product of sin or cos like this then you may be able to use the factor formulae.
    Think you may have to tell me what exercise it is from the book as I'm still having trouble.

    So far I've done...

     \int 2sin(3x)sin(2x) dx

     = \int [cos(5x)-cos(x)]sin(2x) dx

     = \int sin(2x)cos(5x) - sin(2x)cos(x) dx

    But I don't really know what I'm doing
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    (Original post by Philip-flop)
    Think you may have to tell me what exercise it is from the book as I'm still having trouble.

    So far I've done...

     \int 2sin(3x)sin(2x) dx

     = \int [cos(5x)-cos(x)]sin(2x) dx

     = \int sin(2x)cos(5x) - sin(2x)cos(x) dx

    But I don't really know what I'm doing
    Exercise 7E

    These identities aren't as simple to use as most trig identites.

    You have \sin 3x \sin 2x

    And the form that you need in the identity is \sin \frac{A+B}{2}\sin \frac{A-B}{2}

    So you need to work out what A and B have to be. If you think about it long enough you may be able to see this. If not you can say

    \frac{A+B}{2} = 3x and \frac{A-B}{2}=2x

    Then solve these simultaneously to find A and B in terms of x.

    Try that and post your working if you get stuck.
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    (Original post by notnek)
    Exercise 7E

    These identities aren't as simple to use as most trig identites.

    You have \sin 3x \sin 2x

    And the form that you need in the identity is \sin \frac{A+B}{2}\sin \frac{A-B}{2}

    So you need to work out what A and B have to be. If you think about it long enough you may be able to see this. If not you can say

    \frac{A+B}{2} = 3x and \frac{A-B}{2}=2x

    Then solve these simultaneously to find A and B in terms of x.

    Try that and post your working if you get stuck.
    Yeah I actually already worked it out to be A=5x and B=x so then I went ahead to give (using the factor formula) ...

     \int 2sin(3x)sin(2x) dx

     = \int [cos(5x)-cos(x)]sin(2x) dx

    Hope that's right. Then I'm not sure where to go from there
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    (Original post by Philip-flop)
    Yeah I actually already worked it out to be A=5x and B=x so then I went ahead to give (using the factor formula) ...

     \int 2sin(3x)sin(2x) dx

     = \int [cos(5x)-cos(x)]sin(2x) dx

    Hope that's right. Then I'm not sure where to go from there
    Oh right I see you already found A and B. But I'm not sure why you still have the \sin 2x in your integral.

    Rearranging the factor formula gives

    \displaystyle \sin \frac{A+B}{2}\sin \frac{A-B}{2} = -\frac{1}{2}\left(\cos A - \cos B\right)


    So if A=5x and B=x you get

    \displaystyle \sin 3x \sin 2x = -\frac{1}{2}\left (\cos 5x - \cos x\right)

    So your integral is now entirely in terms of cos.
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    (Original post by notnek)
    Oh right I see you already found A and B. But I'm not sure why you still have the \sin 2x in your integral.

    Rearranging the factor formula gives

    \displaystyle \sin \frac{A+B}{2}\sin \frac{A-B}{2} = -\frac{1}{2}\left(\cos A - \cos B\right)


    So if A=5x and B=x you get

    \displaystyle \sin 3x \sin 2x = -\frac{1}{2}\left (\cos 5x - \cos x\right)

    So your integral is now entirely in terms of cos.
    Oops, that's why I couldn't work it out originally! Silly me!

    So...
     \int 2sin(3x)sin(2x) dx

     = -1 \int -2sin(3x)sin(2x) dx

    Using Factor Formula...
     = -1 \int cos(5x) - cos(x) dx

     = -1 [\frac{1}{5} sin(5x) - sin(x)] + c

     = sin(x) - \frac{1}{5} sin(5x) + c
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    (Original post by Philip-flop)
    Oops, that's why I couldn't work it out originally! Silly me!

    So...
     \int 2sin(3x)sin(2x) dx

     = -1 \int -2sin(3x)sin(2x) dx

    Using Factor Formula...
     = -1 \int cos(5x) - cos(x) dx

     = -1 [\frac{1}{5} sin(5x) - sin(x)] + c

     = sin(x) - \frac{1}{5} sin(5x) + c
    That's correct. It's very unlikely that an integral like this will come up in a C4 exam but I can't rule it out.
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    (Original post by notnek)
    That's correct. It's very unlikely that an integral like this will come up in a C4 exam but I can't rule it out.
    Yeah I don't think I'll spend any longer on these types. I haven't come across any in the all the past papers I've done so far so it's not worth learning it inside out. It was fun practice though, and at least I'll have a vague idea if it happens to show up in the real exam!
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    Ok so I'm working through the Edexcel C4 June 2015 paper but am stuck on Question 6.

    Can someone explain how I get from this line to the second line?
    Name:  C4 June 2015 Q6.png
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    (Original post by Philip-flop)
    Ok so I'm working through the Edexcel C4 June 2015 paper but am stuck on Question 6.

    Can someone explain how I get from this line to the second line?
    Name:  C4 June 2015 Q6.png
Views: 15
Size:  5.2 KB
    There are two brackets inside the square root and they have been simplified

    (3-(1+2\sin \theta)) = (2- 2\sin\theta)

    and

    ((1+2\sin \theta)+1) = (2+2\sin \theta)

    Which one of these are you unusure about?
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    (Original post by notnek)
    There are two brackets inside the square root and they have been simplified

    (3-(1+2\sin \theta)) = (2- 2\sin\theta)

    and

    ((1+2\sin \theta)+1) = (2+2\sin \theta)

    Which one of these are you unusure about?
    OMG I'm so stupid!! It's cos I wrote  [3(1+2sin \theta)] instead of  [3-(1+2sin \theta)] such a silly little mistake!!

    Thanks for helping me realise!
 
 
 
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