- #1
robierob12
- 48
- 0
Im doing a differential equation and end up with my last step, which is an integration.
I can't seem to remember how to integrate this one
int. (tanx)/(1-cosx)
hmmm
[(sinx)/(cosx)] / (1-cosx)
u = cosx
du = -sinx dx
throw a negative outside the integral.
du/(u)(1-u)
maby? then use some easy partial frac. decomp.
would that be a correct method to use or is there another way.
Rob
I can't seem to remember how to integrate this one
int. (tanx)/(1-cosx)
hmmm
[(sinx)/(cosx)] / (1-cosx)
u = cosx
du = -sinx dx
throw a negative outside the integral.
du/(u)(1-u)
maby? then use some easy partial frac. decomp.
would that be a correct method to use or is there another way.
Rob