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    (Original post by masryboy94)
    i am abit confused at working out \int sin^2 x\ cos^2 x\ dx
    Try and use trig identities to represent the integral in terms of either sin or cosine to only the 1st power
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    (Original post by browb003)
    Try and use trig identities to represent the integral in terms of either sin or cosine to only the 1st power
    i think i did try that but couldn't get to the answer
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    (Original post by masryboy94)
    i think i did try that but couldn't get to the answer
    What's the answer that's given? I'll have a go at it and see what I get
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    (Original post by browb003)
    What's the answer that's given? I'll have a go at it and see what I get
    \frac{1}{8} x\ - \frac{1}{32} sin4x\ +c
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    (Original post by justinawe)
    He applied the chain rule incorrectly, so if you followed his working you'd get the wrong answer.
    question 34 mixed exercise L integration, the last question, I think there is a mistake , because A should be -27, as you are multiplying 9 with -3
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    (Original post by strawberri)
    I don't NEED an a* but would like one as it would enable me to get a grant if I got two other As..
    unfortunately I got 80ums in C3 which means I have to get 100ums/full marks in C4 to get an A*!!! ermmm....

    (I'm not re-taking C3 as I do French and there are no January A2 exams so everything rests on the summer ones...)
    What's the workload like for French A2? possible to self teach? with AS?
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    for c4 do we need to be able to determine the nature of turning points? for implicit / parametric differentiation?
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    (Original post by otrivine)
    question 34 mixed exercise L integration, the last question, I think there is a mistake , because A should be -27, as you are multiplying 9 with -3
    You can't have a negative area.
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    (Original post by justinawe)
    You can't have a negative area.
    so they ignored the minus sign
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    Is it just me or is C4 so much nicer than C3?! DX
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    (Original post by masryboy94)
    \frac{1}{8} x\ - \frac{1}{32} sin4x\ +c
    You can rewrite the integrand as sin(x)^2 - sin(x)^4

    sin(x)^2 integrates to: x/2 - sin(2x)/4 (+c)

    sin(x)^4 integrates to: 3x/8 - sin(2x)/4 + sin(4x)/32 (+c)

    You get this answer if you rewrite cos(x)^2 = 1 - sin(x)^2
    in the first line. Of course you can rewrite the integrand in terms of cosine as well and get an answer that looks different, but should be equivalent mathematically
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    (Original post by browb003)
    You can rewrite the integrand as sin(x)^2 - sin(x)^4

    sin(x)^2 integrates to: x/2 - sin(2x)/4 (+c)

    sin(x)^4 integrates to: 3x/8 - sin(2x)/4 + sin(4x)/32 (+c)

    You get this answer if you rewrite cos(x)^2 = 1 - sin(x)^2
    in the first line. Of course you can rewrite the integrand in terms of cosine as well and get an answer that looks different, but should be equivalent mathematically
    sorry i lost you on the first part, what is it that you rewrote? and where did you get -sinx^4?
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    (Original post by masryboy94)
    sorry i lost you on the first part, what is it that you rewrote? and where did you get -sinx^4?
     \int sin^2xcos^2x \ dx = \int sin^2x(1 - sin^2x) \ dx = \int sin^2x - sin^4x \ dx

    and then you can use the identity  sin^2x = \frac{1}{2} (1 - cos(2x)) several times in order to get everything to the 1st power
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    (Original post by masryboy94)
    i am abit confused at working out \int sin^2 x\ cos^2 x\ dx
    this might be an easier way:

    use the identities sin^2x=\frac{-1}{2}(cos2x-1) and cos^2x= \frac{1}{2}(cos2x+1)

    therefore the equation can be written as \frac{-1}{2}(cos2x-1) \times \frac{1}{2}(cos2x+1)

    simplify and then expand the brackets above: (\frac{-1}{2}cos2x + \frac {1}{4})(\frac{1}{2}cos2x+\frac{1  }{4}) = \frac{-1}{4}cos^22x - \frac{1}{8}cos2x + \frac{1}{8}cos2x + \frac{1}{16}

    use the identity: cos4x=2cos^22x-1 therefore \frac{1}{2}(cos4x+1)=cos^22x and subbing this into the original equation: \frac{-1}{4}(\frac{1}{2}cos4x+1/2) +\frac{1}{16}

    multiply the brackets above and simplifying: \frac{-1}{8}cos4x-\frac{1}{8}

    now integrate: \displaystyle \int \frac{-1}{8}cos4x-\frac{1}{8} dx
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    sorry but question 30)a) mixed exercise L intrgration,

    I differentiated u=1+2x2

    and got dx=1/4x du

    in the book they did

    x dx = 1/4 du ?

    what?
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    (Original post by otrivine)
    sorry but question 30)a) mixed exercise L intrgration,

    I differentiated u=1+2x2

    and got dx=1/4x du

    in the book they did

    x dx = 1/4 du ?

    what?
     u = 1 + 2x^2

     \frac{du}{dx} = 4x

     \frac{1}{4} du = x \ dx
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    (Original post by browb003)
     \int sin^2xcos^2x \ dx = \int sin^2x(1 - sin^2x) \ dx = \int sin^2x - sin^4x \ dx

    and then you can use the identity  sin^2x = \frac{1}{2} (1 - cos(2x)) several times in order to get everything to the 1st power
    (Original post by gaffer dean)
    this might be an easier way:

    use the identities sin^2x=\frac{-1}{2}(cos2x-1) and cos^2x= \frac{1}{2}(cos2x+1)
    therefore the equation can be written as \frac{-1}{2}(cos2x-1) \times \frac{1}{2}(cos2x+1)

    simplify and then expand the brackets above: (\frac{1}{2}cos2x + \frac {1}{4})(\frac{1}{2}cos2x+\frac{1  }{4}) = \frac{1}{4}cos^22x+ \frac{1}{16}

    use the identity: cos4x=2cos^2x-1 therefore \frac{1}{2}(cos4x+1)=cos^22x and subbing this into the original equation: \frac{-1}{4}(\frac{1}{2}cos4x+1/2) +\frac{1}{16}

    multiply the brackets above and simplifying: \frac{-1}{8}cos4x-\frac{2}{16}

    now integrate: \int \frac{-1}{8}cos4x-\frac{2}{16}
    thank you both !!! now i get it
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    (Original post by browb003)
     u = 1 + 2x^2

     \frac{du}{dx} = 4x

     \frac{1}{4} du = x \ dx
    is it because they do not want to include x so they can cancel things because everything is in terms of u?
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    (Original post by otrivine)
    is it because they do not want to include x so they can cancel things because everything is in terms of u?
    Yeah, as rewriting everything in terms of u when substituting will result (in this case) in an integrand that is much easier to integrate
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    (Original post by masryboy94)
    thank you both !!! now i get it
    It's important to Remember these identities:
    sin^2x=\frac{-1}{2}(cos2x-1), cos^2x= \frac{1}{2}(cos2x+1), cos4x=2cos^22x-1
 
 
 
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