Integration by Parts: Troubleshooting the Formula for x cos 5x dx

[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 please.

 

[courtrigrad]

[tex] \int x\cos 5x = \frac{x}{5}\sin 5x - \frac{1}{5}\int \sin 5x dx [/tex].

[tex] \int x\cos 5x = \frac{x}{5}\sin 5x +\frac{1}{25}\cos 5x [/tex]

So [tex] \int udv = uv-\int vdu [/tex].

[tex] \int \sin 5x = -\frac{\cos 5x}{5} [/tex].

 

[lemurs]

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

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

 

[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

 

[lemurs]

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