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    So I have my C4 exam tomorrow,
    was doing solomon paper L,
    and came across this trig integration:

    integrate: (sin3x)*(cosx)^2
    between the limits pi/4 and 0.

    Someone please shoe me how to get that trig function looking normal and integratable???:eek:
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    ...I have checked out the mark scheme, and it still makes no sense unfortunately.
    They manage to get from (sin3x)*(cosx)^2 to 0.25sin4x+0.5sin2x but i dont know how!!!!
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    \displaystyle \sin 3x=\sin 2x\cos x + \cos 2x\sin x = 2\cos^2 x \sin x + (2\cos^2 x-1)\sin x = \sin x(4\cos^2 x -1)

    If use use this and recognition (you should have products of cos powers and sins) you should be able to do it. Ask if you need more help.
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    (Original post by rmod)
    So I have my C4 exam tomorrow,
    was doing solomon paper L,
    and came across this trig integration:

    integrate: (sin3x)*(cosx)^2
    between the limits pi/4 and 0.

    Someone please shoe me how to get that trig function looking normal and integratable???:eek:
    Which paper, which question?
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    paper L 3b
    ....this is a bit embarrassing but i made a mistake when copying the qn down to work it out... theres no cosx squared only cosx. probs makes things a bit easier...
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    (Original post by rmod)
    paper L 3b
    ....this is a bit embarrassing but i made a mistake when copying the qn down to work it out... theres no cosx squared only cosx. probs makes things a bit easier...
    It doesn't really make it easier - the method is still the same.

    Have you made any progress?
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    ummm not really. It's reached that time of night where my brain starts closing down. I'm just gonna hope that one doesnt come up tomorrow/ I'll be sharper tomorrow and able to get it thank you for the help though- i got the bit you showed me, but get stuck when that extra cosx comes into it.
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    Use the SinA SinB= 2Sin(A+B/2)Cos(A_B/2)

    Formula to turn the product into a sum.
 
 
 
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