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

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

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

I'm absolutely baffled brother, could you guide me through the first step, if you don't mind

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?

SinA +SinB = 2(Sin((A+B)/2)*Cos((A+B)/2))
A+B=8 A-B=6

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

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

is the answer -cosx -1/7cos7x

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