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    How do I integrate 0.5sin^2(2x)cos(x) ? It's an AQA past question and I don't understand the mark scheme!
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    (Original post by pengpengzoom)
    How do I integrate 0.5sin^2(2x)cos(x) ? It's an AQA past question and I don't understand the mark scheme!
    Start with the double angle formula.

    What do you have now?
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    (Original post by pengpengzoom)
    How do I integrate 0.5sin^2(2x)cos(x) ? It's an AQA past question and I don't understand the mark scheme!
    I did this question the other day! Yep, using the double angle formula for sin2x , squaring that all out, giving cos^3(x)sin^2(x) then replacing two of the cos(x)'s ( so cos^2(x)) with 1- sin^2(x), multiplying that out then you can integrate to give it in terms of sin^n(x) 's. If that makes sense.
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    (Original post by ladygabriela97)
    I did this question the other day! Yep, using the double angle formula for sin2x , squaring that all out, giving cos^3(x)sin^2(x) then replacing two of the cos(x)'s ( so cos^2(x)) with 1- sin^2(x), multiplying that out then you can integrate to give it in terms of sin^n(x) 's. If that makes sense.
    what did you get?
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    Expand it to 0.5(4sin^2(x)cos^2(x))cos(x) = 2cos^3(x)sin^2(x)

    Then replace cos^2(x) with 1-sin^2(x) = 2cos(x)(1-sin^2(x))sin^2(x)

    expand = 2cos(x)(sin^2(x)-sin^4(x)) -

    = 2sin^2(x)cos(x) - 2sin^4(x)cos(x)

    Then you should be able to use integration with substitution with u = sin(x), so dx = 1/cos(x) du

    Giving you the integral of 2u^2 - 2u^4

    = 2/3u^3 - 2/5u^5 + c = 2/3sin^3(x) - 2/5sin^5(x) + c

    Hopefully that's right haha
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    (Original post by Zenarthra)
    what did you get?
    Sorry that I didn't reply! I hope your fp3 exam went well though, that last question was awful!
 
 
 
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