arctan integral

**Coco87** · September 27th 2008, 02:21 PM

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

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

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

I set:

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

so:

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

and I get:

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

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

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

Any ideas?

**11rdc11** · September 27th 2008, 02:39 PM

Try subbing

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

**Coco87** · September 27th 2008, 02:45 PM

Originally Posted by 11rdc11

Try subbing

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

Thanks for the reply

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 $\frac{1}{x}$ 's with $\tan{x}$

**Jhevon** · September 27th 2008, 02:52 PM

Originally Posted by Coco87

Thanks for the reply

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 $\frac{1}{x}$ 's with $\tan{x}$

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.

**11rdc11** · September 27th 2008, 02:55 PM

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

Remember to change the other values to theta too

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

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

and use the identity

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

so you get

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

**mr fantastic** · September 27th 2008, 03:04 PM

Originally Posted by Coco87

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

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

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

I set:

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

so:

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

and I get:

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

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

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

Any ideas?

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

Integration by parts leads to a simple calculation.

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

**11rdc11** · September 27th 2008, 03:12 PM

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.

**Coco87** · September 27th 2008, 04:21 PM

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 $x\arctan{\left(\frac{1}{x}\right)}+\left(\frac{1}{ 2}\right)\ln{\left|x^2+1\right|}$ , 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 $\arctan{\left(\frac{1}{x}\right)}$ . 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!

**Jhevon** · September 27th 2008, 04:41 PM

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 $x\arctan{\left(\frac{1}{x}\right)}+\left(\frac{1}{ 2}\right)\ln{\left|x^2+1\right|}$ , 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 $\arctan{\left(\frac{1}{x}\right)}$ . 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

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