ok, so the first question made no sense. it told you to find the answer by writing the answer...but anyway.
for the second question, i would not use that. i would say that:
cos^3x = cos^2x*cosx = (1 - sin^2x)cosx
then use the substitution u = sinx, we'd end up integrating 1 - u^2 ...easy. but in the spirit of being as weird as this question is, let's do it like they said we should.
we are told that
cos(3x) = 4cos^3x - 3cosx
=> 4cos^3x = cos(3x) + 3cosx
=> cos^3x = (1/4)[cos(3x) + 3cosx]
=> int{cos^3x}dx = (1/4)*int{[cos(3x) + 3cosx]}dx
you should be able to find the integral on the right without much problems...if you have problems, don't be afraid to say so
The problem is dealing with cos(3x) not cos^3(x).
For the record:
cos(3x) = cos(2x + x) = cos(2x)cos(x) - sin(2x)sin(x)
= (2cos^2(x) - 1)cos(x) - 2sin(x)cos(x)sin(x)
= 2cos^3(x) - cos(x) - 2sin^2(x)cos(x)
= 2cos^3(x) - cos(x) - 2[1 - cos^2(x)]cos(x)
= 2cos^3(x) - cos(x) - 2cos(x) + 2cos^3(x)
= 4cos^3(x) - 3cos(x)
as advertised.
-Dan
which is what i said.
and in regards to your second earlier post, this is what the question says.
let the formula they have for cos(3x) be f(x)
the question basically said.
by writing cos(3x) as cos(3x) = f(x) show that cos(3x) = f(x)
if that's not redundant i don't know what is
It was one of the proof ones were you had to show/prove it just like what topsquark did...
cos(3x) = cos(2x + x) = cos(2x)cos(x) - sin(2x)sin(x)
= (2cos^2(x) - 1)cos(x) - 2sin(x)cos(x)sin(x)
= 2cos^3(x) - cos(x) - 2sin^2(x)cos(x)
= 2cos^3(x) - cos(x) - 2[1 - cos^2(x)]cos(x)
= 2cos^3(x) - cos(x) - 2cos(x) + 2cos^3(x)
= 4cos^3(x) - 3cos(x)
yeah, i know what topsquark did was correct, i'm just saying that's not how the question was phrased. it should have said something like:
"Prove that cos(3x) = 4cos^3x - 3cosx"
and that's it.
but since no one is seeing my point but me, let's just drop it. you have the correct answers