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    ∫ 2 sin 3x sin 2x dx.

    Multiple methods pls and the simplest way
    thanks !
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    Can be done by parts or use of factor formulae. ie.  \displaystyle \cos A - \cos B = -2\sin \left ( \frac{A+B}{2} \right ) \sin \left ( \frac{A-B}{2} \right ) .
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    This was the markscheme way of doing it: = ∫ (cos x − cos 5x) dx
    idk how they went from the original integral to this....
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    Look at my post above

    (Original post by Someboady)
    This was the markscheme way of doing it: = ∫ (cos x − cos 5x) dx
    idk how they went from the original integral to this....
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    (Original post by B_9710)
    Can be done by parts or use of factor formulae. ie.  \displaystyle \cos A - \cos B = -2\sin \left ( \frac{A+B}{2} \right ) \sin \left ( \frac{A-B}{2} \right ) .
    Ah... why have I never seen this formulae before :/
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    (Original post by B_9710)
    Look at my post above
    Is the formula given (Edexcel) or do I have to memorise that *cries* or can I derive it in an exam?
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    (Original post by Someboady)
    Is the formula given (Edexcel) or do I have to memorise that *cries* or can I derive it in an exam?
    You can derive it but it's easier just to remember it. There's 4 factor formula's you need to know although apparently it's not tested the factor formulae. But best to know I think.
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    (Original post by B_9710)
    You can derive it but it's easier just to remember it. There's 4 factor formula's you need to know although apparently it's not tested the factor formulae. But best to know I think.
    Ah they're given in the formula book yesss, thanks a bunch dude would never have known about them :/ kinda scary
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    I got it using a triple angle formula, a double angle formula then a substitution. It should be pretty simple from here
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    (Original post by Someboady)
    ∫ 2 sin 3x sin 2x dx.

    Multiple methods pls and the simplest way
    thanks !
    Just to add, have a look at your formula booklet (online). For Edexcel I suspect that it is in C3.
 
 
 
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