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How to integrate cos^2 (4x)-c4 Watch

1. (Original post by James Cameron)
Well i don't have a formula book on me right now, would the formula you're talking about be sin(a+b)=sinacosb+sinbcosa
http://filestore.aqa.org.uk/subjects/FORMULAE.PDF
2. (Original post by James Cameron)
Well i don't have a formula book on me right now, would the formula you're talking about be sin(a+b)=sinacosb+sinbcosa
Use sinA + cosB = 2sin((A+B)/2)cos((A-B)/2)

Make simultaneous equations to find A and B and then it's simple to integrate.

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3. I'm absolutely baffled brother, could you guide me through the first step, if you don't mind
4. (Original post by James Cameron)
I'm absolutely baffled brother, could you guide me through the first step, if you don't mind
^^^^^Look above^^^^^^
5. (Original post by powelsmartin)
Use sinA + cosB = 2sin((A+B)/2)cos((A-B)/2)

Make simultaneous equations to find A and B and then it's simple to integrate.

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Sin A + COS B , are you sure?
6. (Original post by zetamcfc)
Sin A + COS B , are you sure?
Not really, that's from memory you should double check the formula booklet, apologies if it's wrong!

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7. Some expensive wine that someone has given to me from dubaii and a box of Swiss chocolates
8. SinA +SinB = 2(Sin((A+B)/2)*Cos((A+B)/2))
A+B=8 A-B=6
9. (Original post by zetamcfc)
SinA +SinB = 2(Sin((A+B)/2)*Cos((A+B)/2))
A+B=8 A-B=6
Right, it's sinA + sinB, sorry! I seem to be failing this thread

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10. (Original post by zetamcfc)
SinA +SinB = 2(Sin((A+B)/2)*Cos((A+B)/2))
A+B=8 A-B=6
Tried this for quite a while and it makes no sense to me , could you please post the first part of this solution so I can figure out the rest
11. (Original post by James Cameron)
Tried this for quite a while and it makes no sense to me , could you please post the first part of this solution so I can figure out the rest
So you get that A = 7x and B = 1x (by solving the simultaneous equations (A+B)/2 = 4x and (A-B)/2 = 6x)
Then you substitute your values for A and B and the expression becomes sin1x + sin7x and you integrate that

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12. (Original post by powelsmartin)
So you get that A = 7x and B = 1x (by solving the simultaneous equations (A+B)/2 = 4x and (A-B)/2 = 6x)
Then you substitute your values for A and B and the expression becomes sin1x + sin7x and you integrate that

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That seems wrong, i was never taught any of that in school
13. is the answer -cosx -1/7cos7x
14. (Original post by James Cameron)
That seems wrong, i was never taught any of that in school
Oh yeah you got to divide the final thing by 2, sorry!

Have you used the identity SinA + SinB = 2sin((A+B)/2)cos((A-B)/2)?

Basically you equate the right hand side with your original expression so (A+B)/2 is 4x and (A-B)/2 is 3x and you find the values of A and B.

Since your original expression doesn't have the 2 in front, you have to halve the final answer (which I accidentally forgot to do).

That's the thing with A2 maths - they expect you to connect different parts of the syllabus even if you haven't done it explicitly before

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15. (Original post by James Cameron)
So if you take into account my previous message outlining my mistake it would be -(1/2)cosx - (1/14)cos7x + c. Never forget the + c.

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16. (Original post by powelsmartin)
Oh yeah you got to divide the final thing by 2, sorry!

Have you used the identity SinA + SinB = 2sin((A+B)/2)cos((A-B)/2)?

Basically you equate the right hand side with your original expression so (A+B)/2 is 4x and (A-B)/2 is 3x and you find the values of A and B.

Since your original expression doesn't have the 2 in front, you have to halve the final answer (which I accidentally forgot to do).

That's the thing with A2 maths - they expect you to connect different parts of the syllabus even if you haven't done it explicitly before

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Thanks man, you've been incredibly helpful, i completely get it now
17. (Original post by James Cameron)
Thanks man, you've been incredibly helpful, i completely get it now
Cheers, I'm glad I could help

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