# Integration by Parts

• Jul 17th 2006, 06:40 PM
c_323_h
Integration by Parts
Hi guys, need some help on these:

1. $\displaystyle \int x (lnx) dx$

2. $\displaystyle \int x(cos5x)dx$
• Jul 17th 2006, 07:14 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
Hi guys, need some help on these:

1. $\displaystyle \int x (lnx) dx$

Let,
$\displaystyle u=\ln x$ and, $\displaystyle v'=x$
Thus,
$\displaystyle u'=1/x$ and, $\displaystyle v=(1/2)x^2$
Thus,
$\displaystyle \frac{1}{2}x^2\ln x-\int \frac{1}{2} xdx$
Thus,
$\displaystyle \frac{1}{2}x^2\ln x-\frac{1}{4}x^2+C$
• Jul 17th 2006, 07:17 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h

2. $\displaystyle \int x(cos5x)dx$

Let,
$\displaystyle u=x$ and $\displaystyle v'=\cos 5x$
Thus,
$\displaystyle u'=1$ and $\displaystyle v=\frac{1}{5}\sin 5x$
Thus,
$\displaystyle \frac{1}{5}x\sin 5x-\int \frac{1}{5}\sin 5xdx$
Thus,
$\displaystyle \frac{1}{5}x\sin 5x+\frac{1}{25}\cos 5x+C$

(I am the quickest integrator in the west!)
• Jul 17th 2006, 07:17 PM
c_323_h
Quote:

Originally Posted by ThePerfectHacker
Let,

$\displaystyle ...-\int \frac{1}{2} xdx$

why isn't it $\displaystyle \frac{1}{2}x^2dx$.
Where did the $\displaystyle x^2$ dissappear to?
• Jul 17th 2006, 07:22 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
why isn't it $\displaystyle \frac{1}{2}x^2dx$.
Where did the $\displaystyle x^2$ dissappear to?

$\displaystyle \int x^n=\frac{x^{n+1}}{n+1}, n\not =-1$
Thus,
$\displaystyle \int xdx=\frac{1}{2}x^2$ but then you need to multiply by $\displaystyle \frac{1}{2}$ thus,
$\displaystyle \frac{1}{2}\cdot \frac{1}{2}x^2=\frac{1}{4}x^2$