#1**+5 **

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!!)

CPhill
Oct 27, 2014

#2**+5 **

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. :(

Melody
Oct 27, 2014

#3**+5 **

Best Answer

I'll trade you one "Integration By Parts" lesson for one "Parametric Equations as they apply to Parabolas" lesson.....

Deal???

CPhill
Oct 27, 2014

#4**0 **

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. :)

Melody
Oct 27, 2014