# Arctangent Integration help

• Feb 28th 2008, 11:27 AM
thegame189
Arctangent Integration help
I need help integrating:

xarctan(x) / ((1+x^2)^2)

I did trig substitutions like this:

u=tan(theta)
du=sec^2(theta) d(theta)

So I ended up with:

Integral of (tan(theta)*arctan(tan(theta))*(sec^2(theta))) / ((1+tan^2(theta))^2)

Which simplifies into:

Integral of:

(Theta)(tan(theta))(sec^2(theta)) / (sec^2(theta))^2

The sec^2(theta)'s cancel so I have:

(theta)(tan(theta))(cos^2(theta)), which simplifies into:

(theta)(sin(theta))(cos(theta))

Is that right, and if so, how do I integrate it?
• Feb 28th 2008, 11:37 AM
TheEmptySet
Try integration by parts with
$\displaystyle u=\tan^{-1}(x)$ and

$\displaystyle dv=\frac{x}{(x^2+1)^2}dx$
• Feb 28th 2008, 11:46 AM
thegame189
I also did that:

u=arctanx
du=1/(x^2+1)

dv=x/(x^2+1)^2
v= -1/(2(x^2+1))

So then I get:

-arctan(x)/(2(x^2+1)) + .5*Integral of 1/(1+x^2)^2

That's UV - Integral of vdu. I don't know what to do from there...
• Feb 28th 2008, 12:01 PM
TheEmptySet
$\displaystyle \int\frac{1}{(1+x^2)^2}dx$

set
$\displaystyle x=tan(\theta)$ then
$\displaystyle dx=sec^2(\theta)d\theta$

so you get

$\displaystyle \int\frac{1}{(1+\tan^2(\theta))^2}sec^2(\theta)d\t heta=$

$\displaystyle \int\frac{sec^2(\theta)}{sec^4(\theta)}d\theta=\in t\cos^2(\theta)d\theta$

I think you can get it from here
• Feb 28th 2008, 12:28 PM
thegame189
I took cos^2(theta) to (1/2)(1+cos(2theta)).

So I now have it down to:

-arctan(x)/(2(x^2+1)) + (1/2)*((1/2)theta + (1/4)sin (2theta))

So, with the subsitution x=tan(theta), theta=arctan(x), so this factors to:

(1/4)arctan(x) + (1/8)sin (2theta))

What do I do with the (1/8)sin(2theta)? It's not good to put arctan(x) in the sin, to have (1/8)sin(2arctan(x)), right?

I put it to 2sin(theta)cos(theta), so it's:

(1/4)arctan(x) + (1/4)sin(theta)cos(theta)

What's the best way to factor the last bit?
• Feb 28th 2008, 02:45 PM
TheEmptySet
since

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

and by defintion

$\displaystyle tan(\theta)=\frac{opposite}{adjacent}$

we can figure the the hypotenuse using the pythagorean theorem...

$\displaystyle h=\sqrt{x^2+1}$

so using the defintion of trig functions...

$\displaystyle \sin(\theta)=\frac{opp}{hyp}=\frac{x}{\sqrt{x^2+1} }$

by a similar argument

$\displaystyle \cos(\theta)=\frac{1}{\sqrt{x^2+1}}$