# Intergration by parts

1. Oct 9, 2006

### lemurs

kay I am having difficulties with this..
Knowing the gerneral formuala of

|uv'=uv- |vu'

i using a nonehomework question i was trying to make sure i had it down pat was having problems..

| x cos 5x dx

but for some reason i don't get the right answer when it done...
If I have u=x, du=1
and
v'=cos 5x dx

v= 1/5 sin5x??? or did i crew up some where I been having trouble here ..

so i can do the substion and all but this stuff is screwing with help plz.

2. Oct 9, 2006

### courtrigrad

$$\int x\cos 5x = \frac{x}{5}\sin 5x - \frac{1}{5}\int \sin 5x dx$$.

$$\int x\cos 5x = \frac{x}{5}\sin 5x +\frac{1}{25}\cos 5x$$

So $$\int udv = uv-\int vdu$$.

$$\int \sin 5x = -\frac{\cos 5x}{5}$$.

Last edited: Oct 9, 2006
3. Oct 9, 2006

### lemurs

kay my major problem is that 1/5 where does it come from.

how does cos 5x dx = 1/5 sin 5x..

4. Oct 9, 2006

### map7s

when you take the derivative of 1/5 sin 5x, you get cos 5x by doing the chain rule...you have to take the derivative of the argument because it is more complex than just an x

5. Oct 9, 2006

### lemurs

t6hanks Maps Think i undersand it now... hopefully the homework will be easier now

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