How can you integrate this:
òsin³ 2x dx

You can expand out the sin³ using the usual formula for sin² (x) in terms of cos(2x), and the formulae for sin(x).cos(y) etc. in terms of sums of sines and cosines.

James.

òsin³ (2x) dx=òsin² (2x).sin(2x) dx
So, is it ok to say: since cos(2x) = 1 - 2sin² (x) ===> cos(4x) = 1 - 2sin² (2x) ===> sin² (2x) = 0.5(1 - cos(4x))

What do we replace the sin(2x) with, is it, sin(2x) = sin(x + x) = 2sin(x)cos(x)

The first part is correct - you need to convert sin³ (2x) to (1/2)sin(x)(1 - cos(4x)) = (1/2)(sin(x) - sin(x)cos(4x)).

Now we know how to integrate sin(x), so that part is OK. Our aim is to get the whole thing into the form

a₁ sin(p₁ x) + ... + a_n sin(p_n x) + b₁ cos(q₁ x) + ... + b_m cos(q_m x)

i.e. as a sum of sines and cosines of multiples of x, with no products sin(x)cos(x) or sin² (x) etc. (Just ignore that formula if the notation just makes things more confusing.)

So, you might know a formula by which you can express sin(A)cos(B) as a sum of sines and cosines, which is exactly what we want. If not, I'll leave you to try and derive this formula from the addition formulae for sin(A ± B), cos(A ± B).

Then apply this formula, and you get something which is easy to integrate.

James.

So, (1/2)sin(2x)(1 - cos(4x)) = (1/2)(sin(x) - sin(x)cos(4x)).

term: sin(x)cos(4x)

Let me try:
cos(4x) = cos(2x + 2x)?
Iam a little lost about how to get it into the form you mentioned.

I know that sin(A ± B) = sin(A)cos(B) ± cos(A)sin(B)
and
cos(A ± B) = cos(A)cos(B) -+ sin(A)sin(B),
but I am not sure how to go about using it in our problem.

Thanks for your help James.

OK, ignore the question for the moment. Just think about the formulae for sin(A ± B), and try and figure out how you can get these to give you a formula for sin(P)cos(Q).

If you're still stuck after playing with it for a while, then I'll give you a hand.

Ok, I'll play with the formula and get back to you.

James,

I've been playing around with it for nearly an hour, but no luck! Can you help?

Oh well, I'm sure it's good for you :-) (I spent over 6 hours on one question of mine not too long ago, and still didn't solve it, so you're not the only one!)

sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
sin(A - B) = sin(A)cos(B) - cos(A)sin(B)

so

sin(A + B) + sin(A - B) = 2.sin(A)cos(B)

so

sin(A)cos(B) = (1/2)(sin(A + B) + sin(A - B))

We can apply this formula in the problem above, and we get

sin³ (2x) = (1/2)(sin(x) - sin(x)cos(4x)) [which is what we had before]
= (1/2)(sin(x) - (1/2)(sin(5x) + sin(-3x)))

which after a bit of tidying up gives us

= (1/2)sin(x) + (1/4)sin(3x) - (1/4)sin(5x)

which you can integrate easily.

Hope that helps,
James.

PS. These formulae are quite useful in a lot of situations. You can also get a formula for sin(A)sin(B) and for cos(A)cos(B) by playing around with the cos(A ± B) formula. Also useful are the 'reverse' versions of these, which express sin(A) + cos(B), etc., as a product - with a bit of playing around you should be able to find these.

Personally, I never bothered learning all these formulae, because I just learned where they come from and then I derive them whenever I need them (which is very rarely nowadays).

Thanks James for that post!
I get it and have now derived the other forms you talk about.