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Edexcel C3,C4 June 2013 Thread

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Original post by masryboy94
i am abit confused at working out sin2x cos2x dx\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
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 :frown:
Original post by masryboy94
i think i did try that but couldn't get to the answer :frown:


What's the answer that's given? I'll have a go at it and see what I get
Original post by browb003
What's the answer that's given? I'll have a go at it and see what I get


18x 132sin4x +c\frac{1}{8} x\ - \frac{1}{32} sin4x\ +c
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
Reply 785
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?
for c4 do we need to be able to determine the nature of turning points? for implicit / parametric differentiation?
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.
Original post by justinawe
You can't have a negative area.


so they ignored the minus sign
Is it just me or is C4 so much nicer than C3?! DX
Original post by masryboy94
18x 132sin4x +c\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
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?
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?


sin2xcos2x dx=sin2x(1sin2x) dx=sin2xsin4x dx \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 sin2x=12(1cos(2x)) sin^2x = \frac{1}{2} (1 - cos(2x)) several times in order to get everything to the 1st power
(edited 10 years ago)
Original post by masryboy94
i am abit confused at working out sin2x cos2x dx\int sin^2 x\ cos^2 x\ dx

this might be an easier way:

use the identities sin2x=12(cos2x1)sin^2x=\frac{-1}{2}(cos2x-1) and cos2x=12(cos2x+1)cos^2x= \frac{1}{2}(cos2x+1)

therefore the equation can be written as 12(cos2x1)×12(cos2x+1)\frac{-1}{2}(cos2x-1) \times \frac{1}{2}(cos2x+1)

simplify and then expand the brackets above: (12cos2x+14)(12cos2x+14)(\frac{-1}{2}cos2x + \frac {1}{4})(\frac{1}{2}cos2x+\frac{1}{4}) = 14cos22x18cos2x+18cos2x+116\frac{-1}{4}cos^22x - \frac{1}{8}cos2x + \frac{1}{8}cos2x + \frac{1}{16}

use the identity: cos4x=2cos22x1cos4x=2cos^22x-1 therefore 12(cos4x+1)=cos22x\frac{1}{2}(cos4x+1)=cos^22x and subbing this into the original equation: 14(12cos4x+1/2)+116\frac{-1}{4}(\frac{1}{2}cos4x+1/2) +\frac{1}{16}

multiply the brackets above and simplifying: 18cos4x18\frac{-1}{8}cos4x-\frac{1}{8}

now integrate: 18cos4x18dx\displaystyle \int \frac{-1}{8}cos4x-\frac{1}{8} dx
(edited 10 years ago)
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?
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+2x2 u = 1 + 2x^2

dudx=4x \frac{du}{dx} = 4x

14du=x dx \frac{1}{4} du = x \ dx
Original post by browb003
sin2xcos2x dx=sin2x(1sin2x) dx=sin2xsin4x dx \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 sin2x=12(1cos(2x)) 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 sin2x=12(cos2x1)sin^2x=\frac{-1}{2}(cos2x-1) and cos2x=12(cos2x+1)cos^2x= \frac{1}{2}(cos2x+1)
therefore the equation can be written as 12(cos2x1)×12(cos2x+1)\frac{-1}{2}(cos2x-1) \times \frac{1}{2}(cos2x+1)

simplify and then expand the brackets above: (12cos2x+14)(12cos2x+14)(\frac{1}{2}cos2x + \frac {1}{4})(\frac{1}{2}cos2x+\frac{1}{4}) = 14cos22x+116\frac{1}{4}cos^22x+ \frac{1}{16}

use the identity: cos4x=2cos2x1cos4x=2cos^2x-1 therefore 12(cos4x+1)=cos22x\frac{1}{2}(cos4x+1)=cos^22x and subbing this into the original equation: 14(12cos4x+1/2)+116\frac{-1}{4}(\frac{1}{2}cos4x+1/2) +\frac{1}{16}

multiply the brackets above and simplifying: 18cos4x216\frac{-1}{8}cos4x-\frac{2}{16}

now integrate: 18cos4x216\int \frac{-1}{8}cos4x-\frac{2}{16}


thank you both !!! now i get it :biggrin:
Original post by browb003
u=1+2x2 u = 1 + 2x^2

dudx=4x \frac{du}{dx} = 4x

14du=x dx \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?
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
Original post by masryboy94
thank you both !!! now i get it :biggrin:

It's important to Remember these identities:
sin2x=12(cos2x1)sin^2x=\frac{-1}{2}(cos2x-1), cos2x=12(cos2x+1)cos^2x= \frac{1}{2}(cos2x+1), cos4x=2cos22x1cos4x=2cos^22x-1
(edited 10 years ago)

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