# Can't get this integral right

1. Feb 5, 2005

### digink

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.

$$\int x cos5x dx$$

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

2. Feb 5, 2005

### dextercioby

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

Daniel.

3. Feb 5, 2005

### digink

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
$$1/5xsin5x - 1/5\int sin5x dx$$
=$$1/5xsin5x - 1/5 cos5x + C$$

4. Feb 5, 2005

### 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

5. Feb 5, 2005

### 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.

6. Feb 5, 2005

### digink

I don't understand what you just said, are you asking a question or trying to answer one? im sorry

7. Feb 5, 2005

### 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.

8. Feb 6, 2005

### digink

wouldnt that be 1/5 cos 5x?

9. Feb 6, 2005

### 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!!

10. Feb 6, 2005

### 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!!

11. Feb 6, 2005

### DR33

sorry, I couldnt 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!!

12. Feb 6, 2005

### dextercioby

With the minus.Sine integrated is minus cosine.

Daniel.