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