Integration By Parts: What am I doing wrong?

**Krooger** · February 3rd 2009, 09:20 AM

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

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

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

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

Where did I go wrong? Thank You

**Opalg** · February 3rd 2009, 10:14 AM

Originally Posted by Krooger

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

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

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

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

Where did I go wrong? Thank You

..

**Krooger** · February 3rd 2009, 02:23 PM

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

**ahawk1** · February 3rd 2009, 03:07 PM

[quote=Krooger;259573] $\int 3xcos(5x)dx$
Let $u = 3x$ so, $du = 3dx$
Let $dv = cos(5x)dx$ so, $v = sin(5x)/5$

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?

**Krooger** · February 3rd 2009, 03:33 PM

Yup, got it

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

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