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What is the antiderivitive of x(0.03+0.01sin(54x))

Guest Oct 27, 2014

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 #3
avatar+80935 
+5

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

Deal???

 

CPhill  Oct 27, 2014
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 #1
avatar+80935 
+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
avatar+91435 
+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
avatar+80935 
+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
avatar+91435 
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

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