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Can't get this integral right

  • Thread starter digink
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  • #1
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I've been doing this for a while and I cant get the same answer as the book. Im just going to give the problem first to see if you guys end up where I did, please explain your work.

[tex]\int x cos5x dx[/tex]

I can solve it just fine using integration by parts but I dont get the right answer
 

Answers and Replies

  • #2
dextercioby
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Why not???It's the only way to do it without getting a headache... :tongue2:

Post your work.

Daniel.
 
  • #3
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ok

this is what im doing. I have u=x dv=cos5x du=1 v=1/5 cos 5x (i forgot what the rule is for a cos with something like 5x, just x cos x?)

with that I get
[tex]1/5xsin5x - 1/5\int sin5x dx[/tex]
=[tex]1/5xsin5x - 1/5 cos5x + C[/tex]
 
  • #4
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btw if it helps any the answer in the back of the book is the same as mine except for the fact that the 1/5 in front of the cos is 1/25
 
  • #5
dextercioby
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And the sign shoud be a plus too... :wink:

Why is it 1/25 and not 1/5.Better put on what condition would the factor be 1/5.What function would you have to integrate??

Daniel.
 
  • #6
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dextercioby said:
And the sign shoud be a plus too... :wink:

Why is it 1/25 and not 1/5.Better put on what condition would the factor be 1/5.What function would you have to integrate??

Daniel.
I don't understand what you just said, are you asking a question or trying to answer one? im :confused: sorry
 
  • #7
dextercioby
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Sorry if i seemed evasive.Didn't mean it.
Compute this
[tex] \int \sin 5x \ dx [/tex]

It's the last integral u had to compute after taking the partial integration initially.

Daniel.
 
  • #8
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dextercioby said:
Sorry if i seemed evasive.Didn't mean it.
Compute this
[tex] \int \sin 5x \ dx [/tex]

It's the last integral u had to compute after taking the partial integration initially.

Daniel.
wouldnt that be 1/5 cos 5x?
 
  • #9
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ok here it is. The answer should be

1/5xsin5x + 1/25 cos5x + C

First you use: u = x
du = dx
dv = cos5x dx
v = 1/5sinx dx

and now you use the integral by parts:

uv - \int vdu = 1/5xsin5x - \int 1/5sin5xdx

u take the 1/5 (constant out the integral) --> 1/5xsin5x - 1/5 \int sin5xdx

= 1/5xsin5x - 1/5 (-1/5cos5x)
= 1/5xsin5x + 1/25cos5x + C

thre you go!!
 
  • #10
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ok here it is. The answer should be

1/5xsin5x + 1/25cos5x + C

First you use: u = x
du = dx
dv = cos5x dx
v = 1/5sinx dx

and now you use the integral by parts:

uv - \int \vdu\ = 1/5xsin5x - \int \1/5sin5x \ dx

u take the 1/5 (constant out the integral) --> 1/5xsin5x - 1/5 \int \sin5x \ dx

= 1/5xsin5x - 1/5 (-1/5cos5x)
= 1/5xsin5x + 1/25cos5x + C

thre you go!!
 
  • #11
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sorry, I couldnt get the symbols working:

[tex]. . . . . . . . . [/tex].

ok here it is. The answer should be

1/5xsin5x + 1/25cos5x + C

First you use: u = x
du = dx
dv = cos5x dx
v = 1/5sinx dx

and now you use the integral by parts:

uv - [tex] \int \vdu\ [/tex] = 1/5xsin5x - [tex] \int \1/5sin5x \ dx [/tex]

u take the 1/5 (constant out the integral) --> 1/5xsin5x - [tex]1/5 \int \sin5x \ dx [/tex]


= 1/5xsin5x - 1/5 (-1/5cos5x)
= 1/5xsin5x + 1/25cos5x + C

thre you go!!
 
  • #12
dextercioby
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digink said:
wouldnt that be 1/5 cos 5x?
With the minus.Sine integrated is minus cosine.

Daniel.
 

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