ok
this is what I am doing. I have u=x dv=cos5x du=1 v=1/5 cos 5x (i forgot what the rule is for a cos with something like 5x, just x cos x?)
with that I get
1/5xsin5x - 1/5\int sin5x dx
=1/5xsin5x - 1/5 cos5x + C
I've been doing this for a while and I can't get the same answer as the book. I am just going to give the problem first to see if you guys end up where I did, please explain your work.
\int x cos5x dx
I can solve it just fine using integration by parts but I don't get the right answer
Hey dexter I am sorry but you sort of lost me at step 2 when you did the integration by parts. Were you using this formula(this is off the top of my head might be slightly different):
\int f(x)g'(x) dx = f(x)g(x) - \int f'(x)g(x) dx ?
Also I am not 100% sure how to implement this at the...
oops for some reason my brain was thinking common denominator, ill try to see if I can solve it with what you just gave.. I figured if their was a mistake it was at that step.
edit: BTW can anyone else tell me if Muzza is right and I was wrong at that step? I am thinking its correct but like...
I was given this problem and the persom whom I received it from told me they couldn't see how it would he solved or even if it had an answer. I think I might have come to a conclusion even thought it might be wrong, what do you guys think
\int tan^2x sec^3x dx
this is what I've been able...
Im only in calc2 right now and even that doesn't help me but I guess I will learn more as I finish the rest of my calc courses.
with x=u-pi x'=1 (if pi is a constant)
sorry i still don't see it lol
wow I feel really stupid to forget that lol, I forgot that when e^x is raised to something like 3x you use the derivative of that times the original e^x function.
thanks :D
Let me put it in a simpler form, just replace u with x something you are probably more common to seeing.
Now the derivative for the ln x = 1/x.
now if you have \int {1/x}dx you know that's the derivative of the ln of x, so you end up with that = ln|x| + C
this is just based of knowing...
Ok I know that the \int e^x = e^x + C
now I don't understand this problem.
\int_0^1 e^{-3x} dx = -(1/3)e^{-3(0)} - -(1/3)e^{-3(1)}
= 1/3(1- e^{-3} )
Where does the 1/3 come from?