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    how would integrate this ...

    ∫ 2 sin 3x sin 2x dx.
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    (Original post by Raysfc16)
    how would integrate this ...

    ∫ 2 sin 3x sin 2x dx.
    Use trigonometry to change the form of the integrand.
    Spoiler:
    Show
    Hint:
    Expand \cos(A-B)-\cos(A+B)
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    (Original post by Raysfc16)
    how would integrate this ...

    ∫ 2 sin 3x sin 2x dx.
    Use your formula booklet. Under the C3 section, it clearly states that -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2} = \cos A - \cos B.

    Use this in the case that A+B = 6x and A-B= 4x, so A=5x and B = x.

    This gives you \displaystyle -\int -2\sin 3x \sin 2x \, \mathrm{d}x = -\int \cos 5x - \cos x \, \mathrm{d}x = \int \cos x - \cos 5x \, \mathrm{d}x
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    Oh, and moved to maths.
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    (Original post by EricPiphany)
    Use trigonometry to change the form of the integrand.
    Spoiler:
    Show
    Hint:
    Expand \cos(A-B)-\cos(A+B)
    (Original post by Zacken)
    Use your formula booklet. Under the C3 section, it clearly states that -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2} = \cos A - \cos B.

    Use this in the case that A+B = 6x and A-B= 4x, so A=5x and B = x.

    This gives you \displaystyle -\int -2\sin 3x \sin 2x \, \mathrm{d}x = -\int \cos 5x - \cos x \, \mathrm{d}x = \int \cos x - \cos 5x \, \mathrm{d}x



    thank you so much guys!
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    (Original post by Raysfc16)
    thank you so much guys!
    Pleasure.
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    It can also be done by parts
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    (Original post by TeeEm)
    It can also be done by parts
    Twice? I'm going to try it
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    (Original post by EricPiphany)
    Twice? I'm going to try it
    Indeed
    Just as quick but not widely known
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    (Original post by EricPiphany)
    Twice? I'm going to try it
    Cyclic IBP's, the kind where you need to re-arrange and solve for I. It's not as quick as spotting the trig simplification, though.
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    (Original post by TeeEm)
    Indeed
    Just as quick but not widely known
    Just as quick?
 
 
 
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