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    Put me out of my misery please!
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    Write \sin^3 2x = \sin 2x \times \sin^2 2x and then use the identity \sin^2 \theta + \cos^2 \theta = 1. You can then integrate the resulting two terms separately; if you get stuck on this let us know.
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    So do I end up with

    Integral of Sin2x multiplied by the integral of (1-cos^(2x))??

    Even so I dont' understand why I would need to convert sin^2x to (1-cos^2x) sorry if I'm missing something because this it what came up when I searched on Google but I don't understand.

    You can multiply integrals together can't you?
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    Ignore that, I get it!!!!

    I think anyway, let me have a crack at a few in the book and I'll come back if I get stuck. Thanks for your help and time
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    Actually, I've got this far: -

    Int (sin2x) - Int (Cos^2x)(Sin2x) <-- How to I integrate this end bit?

    I'm fine with the first bit (sin2x)
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    (Original post by rosschambers1987)
    Actually, I've got this far: -

    Int (sin2x) - Int (Cos^2x)(Sin2x) <-- How to I integrate this end bit?

    I'm fine with the first bit (sin2x)
    What do you get when you differentiate \cos^3 2x?
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    I don't know, my brain isn't working (bangs head on desk).

    Is it just chain rule and I get

    3(cos2x)^2 * 1/2(sin2x)? so it's 3/2(cos^2x)(sin2x) so it would be 2/3(cos^3(2x))?

    So final answer is (-1/2cos2x)-(2/3cos^3(2x))

    I'm nearly there, the answer in the book has a +1/6 instead of my -2/3.
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    Sorry I've got it now, started from scratch again, got the same answer as in the book (I didn't realise I had to chain rule twice on differtiating cos^3(2x).

    Thanks so much.
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    I'm not sure if this is right but I gave it a go...

    Integral of sin^3(2x) dx

    well first you can factorise sin(2x) to get a positive power --->> the integral of sin(x)[(sin^2(2x)]

    then use the trig. identity sin^2(x)+cos^2(x)=1 so the integral of (1-cos^2(2x))sin(x) dx

    now use substitution, u=cos(2x) the derivative of that is -2sin(2x)

    there dx=du/-2sin(2x)

    so the integral of (1-u^2)sin(2x) du/-2sin(2x)

    you can cancel out the sin(2x) function and factorise out the negative half to get

    -1/2 integral 1-u^2 du

    then just integrate to get -1/2(u-u^3/3)+c

    then you get -1/2cos(2x)- 1/3cos^3(2x)+c

    I may be wrong so check out this video on youtube and see if you can follow the steps
    http://www.youtube.com/watch?v=mLvRpWBhlTA
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    (Original post by A15hat)
    I'm not sure if this is right but I gave it a go...

    Integral of sin^3(2x) dx

    well first you can factorise sin(2x) to get a positive power --->> the integral of sin(x)[(sin^2(2x)]

    then use the trig. identity sin^2(x)+cos^2(x)=1 so the integral of (1-cos^2(2x))sin(x) dx

    now use substitution, u=cos(2x) the derivative of that is -2sin(2x)

    there dx=du/-2sin(2x)

    so the integral of (1-u^2)sin(2x) du/-2sin(2x)

    you can cancel out the sin(2x) function and factorise out the negative half to get

    -1/2 integral 1-u^2 du

    then just integrate to get -1/2(u-u^3/3)+c

    then you get -1/2cos(2x)- 1/3cos^3(2x)+c

    I may be wrong so check out this video on youtube and see if you can follow the steps
    http://www.youtube.com/watch?v=mLvRpWBhlTA
    It is almost right.
    At the last row the factor of cosine function will be (-1/2)(-1/3)=1/6
 
 
 
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