# Integration By Parts: What am I doing wrong?

• Feb 3rd 2009, 09:20 AM
Krooger
Integration By Parts: What am I doing wrong?
$\displaystyle \int 3xcos(5x)dx$
Let $\displaystyle u = 3x$ so, $\displaystyle du = 3dx$
Let $\displaystyle dv = cos(5x)dx$ so, $\displaystyle v = sin(5x)/5$

Since, $\displaystyle \int udv = uv - \int duv$

$\displaystyle \int 3xcos(5x)dx = 3xsin(5x)/5 - (1/15)\int sin(5x)dx$
$\displaystyle = 3xsin(5x)/5 - (1/15)[-cos(5x)/5]$
$\displaystyle = (3/5)xsin(5x) + (1/75)cos(5x)$

...but this is wrong.
the answer is: $\displaystyle (3/25)(cos(5x)+5xsin(5x))$

Where did I go wrong? Thank You
• Feb 3rd 2009, 10:14 AM
Opalg
Quote:

Originally Posted by Krooger
$\displaystyle \int 3xcos(5x)dx$
Let $\displaystyle u = 3x$ so, $\displaystyle du = 3dx$
Let $\displaystyle dv = cos(5x)dx$ so, $\displaystyle v = sin(5x)/5$

Since, $\displaystyle \int udv = uv - \int duv$

$\displaystyle \int 3xcos(5x)dx = 3xsin(5x)/5 - ({\color{red}3/5})\int sin(5x)dx$ (You divided by 3 instead of multiplying by 3.)
$\displaystyle = 3xsin(5x)/5 - (1/15)[-cos(5x)/5]$
$\displaystyle = (3/5)xsin(5x) + (1/75)cos(5x)$

...but this is wrong.
the answer is: $\displaystyle (3/25)(cos(5x)+5xsin(5x))$

Where did I go wrong? Thank You

..
• Feb 3rd 2009, 02:23 PM
Krooger
Ah I see, I was thinking you had to isolate dx in order to sub it in making my result du/3 instead of 3du.

Thank You
• Feb 3rd 2009, 03:07 PM
ahawk1
[quote=Krooger;259573]$\displaystyle \int 3xcos(5x)dx$
Let $\displaystyle u = 3x$ so, $\displaystyle du = 3dx$
Let $\displaystyle dv = cos(5x)dx$ so, $\displaystyle v = sin(5x)/5$

alright so u do uv-integral(vdu)
so,
3xsin(5x)/5-integral(sin(5x)/5*3dx
3xsin(5x)/5+3cos(5x)/25
do u understand how i got that?
or need me to explain further?
• Feb 3rd 2009, 03:33 PM
Krooger
Yup, got it (Rofl) I was thrown off from the fact that you can just use du, you dont have to isolate and replace dx. Because of that I was using du/3 not 3du.

Thank You again