Question

i dont know this one... integrate arctan (x/3) dx

Answer 1

Think you can say x = Tan y = sin y/cosy and use the f'/f integration.

Answer 2

Or x/3 = Tan y maybe.

Answer 3

Or maybe the trick with multiplying by 1.

Answer 4

i am working with arc tan

Answer 5

a little advanced...are you at integration by parts yet?

Answer 6

Firstly, I would let p=x/3
dp=(1/3)dx
3dp=dx
This makes your integral:
\[3 \int\limits \arctan(p)dp\]
Now, you must use integration by parts.
You let u=arctan(p) (something easy to differentiate)
and dp=dv (something easy to integrate)
So:
\[du=\frac{dp}{1+p^2}\]
And v=p
Now use the formula:
\[uv-\int\limits v du \rightarrow 3p \arctan(p)-3\int\limits \frac{p}{p^2+1}dp\]
For the second integral use the substitution s=p^2+1 => ds=2p dp => (1/2)ds=dp
Your integral now becomes:
\[3p \arctan(p)-\frac{3}{2} \int\limits \frac{ds}{s}=3p \arctan(p)-\frac{3}{2}\ln|s|+C\]
Plugging back in:
\[3p \arctan(p)-\frac{3}{2}\ln|p^2+1|+C\]
Then p.
\[x\arctan(\frac{x}{3})-\frac{3}{2}\ln|(\frac{x}{3})^2+1|+C\]
If you look it up on wolfram they factor out a stuff inside the ln and let it be absorbed by C, that is not really necessary imo.

Answer 7

ok,, perfect.. let me try it...