Recent content by digink

wouldnt that be 1/5 cos 5x?

I don't understand what you just said, are you asking a question or trying to answer one? I am :confused: sorry

btw if it helps any the answer in the back of the book is the same as mine except for the fact that the 1/5 in front of the cos is 1/25

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

that to me looks like it can also work, I am feeling really lazy right now or i'd solve it, ill try it tomorrow. Thanks for the assistance guys.

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

Hey Daniel could you please show me how that'd be solved? I am just curious. I have an idea but I don't see it. Thanks.

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

anyone?? have to study for a test any help would be greatly appreciated. I just want to know where the constant -1/3 came from.

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?