1. ## arctan integral

Hey,
I am on a problem that involves integrating a arctan:

$\displaystyle \int{\arctan{\left(\frac{1}{x}\right)}}dx$

I think what is throwing me off is the $\displaystyle \frac{1}{x}$

I set:

$\displaystyle u=\arctan{\left(\frac{1}{x}\right)}$
$\displaystyle dv=dx$

so:

$\displaystyle du=\frac{1}{1+\left(\frac{1}{x}\right)^2}$
$\displaystyle v=x$

and I get:

$\displaystyle x\arctan{\left(\frac{1}{x}\right)}-\int{\frac{x}{1+\left(\frac{1}{x}\right)^2}}dx$

I think $\displaystyle \int{\frac{x}{1+\left(\frac{1}{x}\right)^2}}dx$ is the problem for me. No matter how I work it, I can't seem to get the answer (maple):

$\displaystyle x\arctan{\left(\frac{1}{x}\right)}+\frac{1}{2}\ln{ \left(1+\frac{1}{x^2}\right)}-\ln{\left(\frac{1}{x}\right)}$

Any ideas?

2. Try subbing

$\displaystyle \frac{1}{x} = \tan\theta$

3. Originally Posted by 11rdc11
Try subbing

$\displaystyle \frac{1}{x} = \tan\theta$

I'm not sure what you mean by subbing though

I'm in Calc 2 on Integrating by Parts; not sure if I have been tought that technique yet, or maybe I've totally missed that part from past courses

By subbing, do you mean replace all of the $\displaystyle \frac{1}{x}$'s with $\displaystyle \tan{x}$

4. Originally Posted by Coco87

I'm not sure what you mean by subbing though

I'm in Calc 2 on Integrating by Parts; not sure if I have been tought that technique yet, or maybe I've totally missed that part from past courses

By subbing, do you mean replace all of the $\displaystyle \frac{1}{x}$'s with $\displaystyle \tan{x}$
he is talking about a trig substitution. chances are you learned that technique before learning the integration by parts technique

anyway, you took the derivative of arctan(1/x) incorrectly. you need to use the chain rule. chances are that's what's messing you up.

5. Yep put $\displaystyle \tan\theta = \frac{1}{x}$

Remember to change the other values to theta too

$\displaystyle x = \frac{1}{\tan\theta}$

$\displaystyle dx = -\csc^2\theta~d\theta$

and use the identity

$\displaystyle 1 + \tan^2\theta = \sec^2\theta$

so you get

$\displaystyle \int \frac{-\csc^2\theta}{\tan\theta\sec^2\theta}~d\theta$

6. Originally Posted by Coco87
Hey,
I am on a problem that involves integrating a arctan:

$\displaystyle \int{\arctan{\left(\frac{1}{x}\right)}}dx$

I think what is throwing me off is the $\displaystyle \frac{1}{x}$

I set:

$\displaystyle u=\arctan{\left(\frac{1}{x}\right)}$
$\displaystyle dv=dx$

so:

$\displaystyle du=\frac{1}{1+\left(\frac{1}{x}\right)^2}$
$\displaystyle v=x$

and I get:

$\displaystyle x\arctan{\left(\frac{1}{x}\right)}-\int{\frac{x}{1+\left(\frac{1}{x}\right)^2}}dx$

I think $\displaystyle \int{\frac{x}{1+\left(\frac{1}{x}\right)^2}}dx$ is the problem for me. No matter how I work it, I can't seem to get the answer (maple):

$\displaystyle x\arctan{\left(\frac{1}{x}\right)}+\frac{1}{2}\ln{ \left(1+\frac{1}{x^2}\right)}-\ln{\left(\frac{1}{x}\right)}$

Any ideas?
To get du you have to use the chain rule: $\displaystyle du = \frac{1}{1 + \left(\frac{1}{x}\right)^2} \cdot {\color{red}\left( - \frac{1}{x^2}\right) } dx = \frac{-1}{x^2 + 1} \, dx$.

Integration by parts leads to a simple calculation.

Note: All the ln stuff Maple gives you is equivalent to $\displaystyle \frac{1}{2} \ln (x^2 + 1) \, ....$

7. Originally Posted by Jhevon
he is talking about a trig substitution. chances are you learned that technique before learning the integration by parts technique

anyway, you took the derivative of arctan(1/x) incorrectly. you need to use the chain rule. chances are that's what's messing you up.
Thanks i forgot to check his work.

8. Sorry about the late response here, but I've been messing with the problem

Thank you all for your help!

I actually looked up trig substitution, and it's 2 sections after this one (hopefully it makes dealing with this stuff easier ). I also need to get use to maple, I got $\displaystyle x\arctan{\left(\frac{1}{x}\right)}+\left(\frac{1}{ 2}\right)\ln{\left|x^2+1\right|}$, and I kept on comparing it to the answer maple got, and it kept showing me false. I finally just did the derivative of my answer, and then simplified it and got $\displaystyle \arctan{\left(\frac{1}{x}\right)}$. So then I played with the answer maple gave me and simplified it symbolically, and it gave me the same answer as I got.

I guess I should RTFM

lol, thanks again guys!

9. Originally Posted by Coco87
Sorry about the late response here, but I've been messing with the problem

Thank you all for your help!

I actually looked up trig substitution, and it's 2 sections after this one (hopefully it makes dealing with this stuff easier ). I also need to get use to maple, I got $\displaystyle x\arctan{\left(\frac{1}{x}\right)}+\left(\frac{1}{ 2}\right)\ln{\left|x^2+1\right|}$, and I kept on comparing it to the answer maple got, and it kept showing me false. I finally just did the derivative of my answer, and then simplified it and got $\displaystyle \arctan{\left(\frac{1}{x}\right)}$. So then I played with the answer maple gave me and simplified it symbolically, and it gave me the same answer as I got.

I guess I should RTFM

lol, thanks again guys!
RTFM?

anyway, Mr F told you that answer was equivalent