3
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Problem:

Show that $$\int_{0}^{\pi / 2} \ln\left(\tan x - \sqrt{2 \tan x} + 1\right)\,\mathrm{d}x = 0 $$ I'd like to use, if possible, only single-variable Calculus methods, and it does not include series manipulations. Only substitutions, IBP, partial fractions and so on. Thanks.

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  • $\begingroup$ suspect this is false, but might work if you had $2 \sqrt {\tan x} $ instead of $ \sqrt {2 \tan x}. $ In any case, where did you get this? $\endgroup$ – Will Jagy Sep 5 '14 at 2:01
  • $\begingroup$ Wolfram and Maple say this is true. I need to prove this to compute $$ \int_{-\infty}^{\infty} \frac{\left(x^2 - 1\right)\arctan(x^2)}{1 + x^4}\,\mathrm{d}x $$. $\endgroup$ – user149844 Sep 5 '14 at 2:09
  • $\begingroup$ @user149844: The value to your integral in the comment is $ \ln ( 2 ) \sqrt{2}\pi $ $\endgroup$ – Mhenni Benghorbal Sep 5 '14 at 2:45
  • $\begingroup$ But how could I deduce this value? In an equivalent manner, one could show that $$\int_{0}^{\pi /2} \ln \left(\tan x + \sqrt{2 \tan x} + 1\right)\,\mathrm{d}x = -2\int_{0}^{\pi /2}\ln\left( \cos x\right)\,\mathrm{d}x = -2\int_{0}^{\pi /2} \ln\left(\sin x\right)\,\mathrm{d}x = \pi \ln 2 $$. Just add and subtract $ \ln\left(\tan x +\sqrt{2 \tan x} + 1\right) $ to the integrand and use logarithms properties, but I do not know if this makes the problem someway easier. $\endgroup$ – user149844 Sep 5 '14 at 4:54

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