What is the antiderivitive of x(0.03+0.01sin(54x))
Mmmmm...this looks like a "parts" problem
So we have
∫.03x dx + ∫.01xsin(54x) dx =
.03∫x dx + .01∫x sin(54x) dx
The first is easy...it's just .03[x^2/2} + C = .015x^2 + C
The second requires a bit more doing ....let's forget the .01 out in front....we'll tack it on at the end....
let u = x and du = dx
let v = -(1/54)cos(54x) and dv = sin(54x)dx
So we have
∫u dv =
u*v - ∫v du =
-(x)(1/54)[cos(54x) - ∫-(1/54)cos(54x) dx and......evaluating the second integral, we have
+(1/54)∫cos(54x) dx = [(1/54)^2][sin(54x)] + C
So we end up with (hopefully!!)....
[.015x^2 + C ] + .01 [ -(1/54)(x)[cos(54x) + [(1/54)^2] sin(54x) + C]= (final answer) =
[.015x^2) + (1/100)(1/2916)sin(54x) - (1/5400)(x)cos(54x)] + C
(Yep...I verified this with WolframAlpha....it's definitely a strange one!!)
Thanks Chris :)
This is yet another that I would really like to find the time to examine properly.
there just is not enough hours in a day. :(
I'll trade you one "Integration By Parts" lesson for one "Parametric Equations as they apply to Parabolas" lesson.....
Yes, actually I can probably do integration by parts. I need lessons in matrices and vectors.
Yes we should do the swap. It is all a matter of time. You know when Christmas comes this places slows to a snails pace. We should put effort into teaching each other then. I shall look forward to it. :)