Hi guys, need some help on these:

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

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

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- Jul 17th 2006, 06:40 PMc_323_hIntegration 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 PMThePerfectHackerQuote:

Originally Posted by**c_323_h**

$\displaystyle u=\ln x$ 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 PMThePerfectHackerQuote:

Originally Posted by**c_323_h**

$\displaystyle u=x$ 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 PMc_323_hQuote:

Originally Posted by**ThePerfectHacker**

Where did the $\displaystyle x^2$ dissappear to? - Jul 17th 2006, 07:22 PMThePerfectHackerQuote:

Originally Posted by**c_323_h**

Thus,

$\displaystyle \int xdx=\frac{1}{2}x^2$ but then you need to multiply by $\displaystyle \frac{1}{2}$ thus,

$\displaystyle \frac{1}{2}\cdot \frac{1}{2}x^2=\frac{1}{4}x^2$