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

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- Mar 12th 2010, 11:18 AMKrizalidAn arctan integral
Evaluate $\displaystyle \int_{0}^{1}{\frac{\arctan \left( \sqrt{2+x^{2}} \right)}{\left( 1+x^{2} \right)\sqrt{2+x^{2}}}\,dx}.$

- Mar 14th 2010, 01:28 PMKrizalid
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.

- Mar 15th 2010, 06:26 AMDeadstar
Posting in thread so I remember about it. I think it has to do with...

$\displaystyle \frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}$.

Then find an expression for arctan(x)/x maybe and use integration by parts. I'll look at it later. - Mar 16th 2010, 06:15 AMDeadstar
Will just post any working as I go along...

__Spoiler__: - Mar 17th 2010, 11:34 PMsimplependulum
I make use of (again) 'magic differentiation' to solve this problem .

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

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

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

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

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

The integral

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

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

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

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

Consider $\displaystyle 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 !

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

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

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

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

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

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

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

$\displaystyle = \frac{5 \pi^2}{96} $ - Mar 18th 2010, 04:44 PMKrizalid
okay, my solution uses the fact that

$\displaystyle \frac{\arctan \left( \frac{1}{\sqrt{2+x^{2}}} \right)}{\sqrt{2+x^{2}}}=\int_{0}^{1}{\frac{dt}{2+ x^{2}+t^{2}}}.$ - Apr 17th 2010, 08:21 AMRandom Variable
- Apr 17th 2010, 10:18 AMKrizalid
as a double one.

- Apr 17th 2010, 10:43 AMRandom Variable
- Apr 23rd 2010, 10:08 AMKrizalid
sorry for the late answer, actually, it holds for a better way.

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