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Quick C4 Integration Question



How did they get from the first line to the second line? :confused:
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Original post by creativebuzz


How did they get from the first line to the second line? :confused:


factor out -3a^2, product to sum formula with sin(3t)sin(t)

Posted from TSR Mobile
Original post by Arithmeticae
factor out -3a^2, product to sum formula with sin(3t)sin(t)

Posted from TSR Mobile


What are you referring to exactly when you say the "product to sum formula"?
Original post by creativebuzz


How did they get from the first line to the second line? :confused:

∫asin⁡(t)(−3asin⁡(3t))dt=−3a2∫sin⁡(t)sin⁡(3t)dt=−3a22∫(2sin⁡(−t)sin⁡(3t))dt\displaystyle \int a \sin(t) \left(-3a \sin(3t) \right) dt = -3a^2 \int \sin(t) \sin(3t) dt = -\dfrac{3a^2}{2} \int (2 \sin(-t) \sin(3t)) dt.

Now, notice that sin⁡(−t)sin⁡(3t)=sin⁡(t−2t)sin⁡(t+2t)\sin(-t) \sin(3t) = \sin(t-2t) \sin(t+2t), and use the addition formula 2sin⁡(a+b)sin⁡(a−b)=cos⁡(2b)−cos⁡(2a)2 \sin(a+b) \sin(a-b) = \cos(2b) - \cos(2a).

EDIT: That addition formula can be seen from differentiating the RHS, cos⁡(2x)−cos⁡(2a)\cos(2x) - \cos(2a) with respect to x, noticing that it's −2sin⁡(2x)-2 \sin(2x).
Differentiating the LHS instead, 2sin⁡(a+x)sin⁡(a−x)2 \sin(a+x) \sin(a-x), gives −2sin⁡(a+x)cos⁡(a−x)+2cos⁡(a+x)sin⁡(a−x)=2(cos⁡(a+x)sin⁡(a−x)−sin⁡(a+x)cos⁡(a−x))- 2 \sin(a+x) \cos(a-x) + 2 \cos(a+x) \sin(a-x) = 2 (\cos(a+x) \sin(a-x) - \sin(a+x) \cos(a-x)).
That is 2sin⁡((a−x)−(a+x))=−2sin⁡(2x)2 \sin((a-x)-(a+x)) = -2 \sin(2x) by addition formula.

Hence the LHS and RHS differentiate to the same thing, so they differ by adding a constant. We just need to check that the constant is 0.

But it is: evaluate at x=a to get 0 on the RHS and 0 on the LHS.
(edited 9 years ago)
Reply 4
Avatar for Rkai01
Rkai01
6
Original post by Smaug123
∫asin⁡(t)(−3asin⁡(3t))dt=−3a2∫sin⁡(t)sin⁡(3t)dt=−3a22∫(2sin⁡(−t)sin⁡(3t))dt\displaystyle \int a \sin(t) \left(-3a \sin(3t) \right) dt = -3a^2 \int \sin(t) \sin(3t) dt = -\dfrac{3a^2}{2} \int (2 \sin(-t) \sin(3t)) dt.

Now, notice that sin⁡(−t)sin⁡(3t)=sin⁡(t−2t)sin⁡(t+2t)\sin(-t) \sin(3t) = \sin(t-2t) \sin(t+2t), and use the addition formula 2sin⁡(a+b)sin⁡(a−b)=cos⁡(2b)−cos⁡(2a)2 \sin(a+b) \sin(a-b) = \cos(2b) - \cos(2a).


You don't even have to do that. You can just use the factor formula which is given.
Original post by Rkai01
You don't even have to do that. You can just use the factor formula which is given.

Fair enough. I never bothered to learn the factor formulae. I only ever remember the easy ones.
Reply 6
Avatar for Rkai01
Rkai01
6
Original post by Smaug123
Fair enough. I never bothered to learn the factor formulae. I only ever remember the easy ones.


That is the easier way tho my friend :/
Original post by Rkai01
That is the easier way tho my friend :/

If you insist :smile: I've had enough practice at A-level that all these methods are pretty much equally easy, so I went for the method that seemed most natural to me. Whatever works fastest in the exam, of course, is the best method :smile:
Reply 8
Avatar for davros
davros
16
Original post by Smaug123
Fair enough. I never bothered to learn the factor formulae. I only ever remember the easy ones.


I've never learnt the factor formulae either, but I believe they're given in A level formula booklets, so it's the expected way to go here :smile:
Original post by Arithmeticae
factor out -3a^2, product to sum formula with sin(3t)sin(t)

Posted from TSR Mobile


Are you doing c4 too brah?

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