# An arctan integral

• March 12th 2010, 12:18 PM
Krizalid
An arctan integral
Evaluate $\int_{0}^{1}{\frac{\arctan \left( \sqrt{2+x^{2}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx}.$
• March 14th 2010, 02:28 PM
Krizalid
okay the problem at first glance it's hard, but i don't want to give any hint yet, so i'll wait a couple of days and i'll post a solution.
• March 15th 2010, 07:26 AM
Posting in thread so I remember about it. I think it has to do with...
$\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}$.

Then find an expression for arctan(x)/x maybe and use integration by parts. I'll look at it later.
• March 16th 2010, 07:15 AM
Will just post any working as I go along...

Spoiler:

Let $y = \sqrt{2 + x^2}$, $dy = \frac{x}{\sqrt{2 + x^2}}
dx$

Then for limits we have $0 \to \sqrt{2}$ and $1 \to \sqrt{3}$

So our new integral is $\int_{\sqrt{2}}^{\sqrt{3}} \frac{\arctan(y)}{(y^2 - 1)(\sqrt{y^2 - 2})} dy$
• March 18th 2010, 12:34 AM
simplependulum
I make use of (again) 'magic differentiation' to solve this problem .

Consider $\int_0^1 \frac{dx}{(x^2+1)\sqrt{x^2+2} }$

Sub. $x = \frac{1}{t}$ , it becomes

$\int_1^{\infty} \frac{t~dt}{(t^2+1)\sqrt{1+2t^2} }$

sub $1+2t^2 = u^2$ .... After a few steps , we can obtain

$\int_0^1 \frac{dx}{(x^2+1)\sqrt{x^2+2} } = \frac{\pi}{6}$

The integral

$\int_{0}^{1}{\frac{\arctan \left( \sqrt{2+x^{2}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx}$

$= \int_{0}^{1}{\frac{\frac{\pi}{2} - \arctan \left( \frac{1}{\sqrt{2+x^{2}}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx}$

$= \frac{\pi}{2} \int_0^1 \frac{dx}{(x^2+1)\sqrt{x^2+2} } - \int_{0}^{1}{\frac{\arctan \left( \frac{1}{\sqrt{2+x^{2}}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx}$

$= \frac{\pi^2}{12} - \int_{0}^{1}{\frac{\arctan \left( \frac{1}{\sqrt{2+x^{2}}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx}$

Consider $I(a) = \int_{0}^{1}{\frac{\arctan \left( \frac{a}{\sqrt{2+x^{2}}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx}$

Magic Differentiation !

$I'(a) = \int_0^1 \frac{dx}{(x^2+1)(x^2 + a^2 + 2 )}$

$= \frac{1}{1 + a^2 } \int_0^1 \left[ \frac{1}{1 + x^2 } - \frac{1}{x^2 + a^2+2} \right]~dx$

$= \frac{\pi}{4}\cdot \frac{1}{1+a^2} - \int_{0}^{1}{\frac{\arctan \left( \frac{1}{\sqrt{2+a^{2}}} \right)}{\left( 1+a^{2} \right)\sqrt{2+a^{2}}}\,dx}$

So $\int_{0}^{1}{\frac{\arctan \left( \frac{1}{\sqrt{2+x^{2}}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx} = I(1) = I(1)-I(0)$

$= \frac{\pi^2}{16}- I(1)$

$= \frac{\pi^2}{32}$

The answer to the problem is $\frac{\pi^2}{12} - \frac{\pi^2}{32}$

$= \frac{5 \pi^2}{96}$
• March 18th 2010, 05:44 PM
Krizalid
okay, my solution uses the fact that

$\frac{\arctan \left( \frac{1}{\sqrt{2+x^{2}}} \right)}{\sqrt{2+x^{2}}}=\int_{0}^{1}{\frac{dt}{2+ x^{2}+t^{2}}}.$
• April 17th 2010, 09:21 AM
Random Variable
Quote:

Originally Posted by Krizalid
okay, my solution uses the fact that

$\frac{\arctan \left( \frac{1}{\sqrt{2+x^{2}}} \right)}{\sqrt{2+x^{2}}}=\int_{0}^{1}{\frac{dt}{2+ x^{2}+t^{2}}}.$

Did you write the original integral as a triple integral and then change the order of integration?
• April 17th 2010, 11:18 AM
Krizalid
as a double one.
• April 17th 2010, 11:43 AM
Random Variable
Quote:

Originally Posted by Krizalid
as a double one.

$\int^{1}_{0} \int^{\infty}_{1} \frac{dt \ dx}{(1+x^{2})(2+x^{2}+t^{2})}$ ?
• April 23rd 2010, 11:08 AM
Krizalid
sorry for the late answer, actually, it holds for a better way.

try to find it, my time is bounded by now.