Math Help - [SOLVED] Antiderivatives

1. [SOLVED] Antiderivatives

I'm having trouble integrating this:
f'(x)= (4/3)cos (1/3 x) - 4sin(2/3 x)

I feel like I should maybe use the chain rule (?)...Any help would be appreciated.
Thank you!

2. Originally Posted by yzobel
I'm having trouble integrating this:
f'(x)= (4/3)cos (1/3 x) - 4sin(2/3 x)

I feel like I should maybe use the chain rule (?)...Any help would be appreciated.
Thank you!
you're integrating... there's no chain rule. you would integrate (using substitution if you need to)

to integrate (4/3)cos[(1/3)x] you can use u = (1/3)x

to integrate 4sin[(2/3)x] you can use t = (2/3)x

3. Substitution is, essentially, the "reverse" of the chain rule.

Let u= x/3 in the first integral and let v= 2x/3 in the second.

(I am assuming that "1/3 x" means "(1/3)x", not "1/(3x)".)

4. Ok, Get it now. Thank you!

5. Originally Posted by yzobel
We haven't covered substitution yet so I'm not really following, but I appreciate the help anyways. Thank you
well, in that case, you have to use a little insight. try to think of what you would differentiate to get each term that you have.

example, say i was asked for the antiderivative of cos(2x), i would reason as follows:

"Hmm, well, I know i have to differentiate sine to get cosine, and the angle doesn't change, so this must come from something like sin(2x). but there's a problem, if i differentiate sin(2x) i would get 2cos(2x), and i don't have that 2. No problem, I will just divide by 2 then. so the antidervative is (1/2)sin(2x) + C. when i differentiate this, i get cos(2x)"

got it? substitution in this case seeks to formalize this kind of reasoning, and gets rid of some of the guesswork, but you can do it without it here. HallsOfIvy is correct though. the point of substitution is to reverse the chain rule. (it was the chain rule that demanded we put the 2 in front of the cos(2x), dividing by (1/2) canceled it). but there are cases that will be more complicated than this one. look forward to it!

now, apply the idea here to your problem