Troubleshooting Integration: Solving \int x cos5x dx

[digink]

I've been doing this for a while and I can't get the same answer as the book. I am just going to give the problem first to see if you guys end up where I did, please explain your work.

\int x cos5x dx

I can solve it just fine using integration by parts but I don't get the right answer

 

[dextercioby]

Why not?It's the only way to do it without getting a headache...

Post your work.

Daniel.

 

[digink]

ok

this is what I am 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
1/5xsin5x - 1/5\int sin5x dx
=1/5xsin5x - 1/5 cos5x + C

 

[digink]

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

 

[dextercioby]

And the sign shoud be a plus too...

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.

 

[digink]

And the sign shoud be a plus too...

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? I am sorry

 

[dextercioby]

Sorry if i seemed evasive.Didn't mean it.
Compute this
\int \sin 5x \ dx

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

Daniel.

 

[digink]

Sorry if i seemed evasive.Didn't mean it.
Compute this
\int \sin 5x \ dx

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

Daniel.

wouldnt that be 1/5 cos 5x?

 

[DR33]

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!

 

[DR33]

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!

 

[DR33]

sorry, I couldn't get the symbols working:

. . . . . . . . ..

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!

 

[dextercioby]

wouldnt that be 1/5 cos 5x?

With the minus.Sine integrated is minus cosine.

Daniel.