# How can I show that the integral equals zero?

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.

• 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? – Will Jagy Sep 5 '14 at 2:01
• Wolfram and Maple say this is true. I need to prove this to compute $$\int_{-\infty}^{\infty} \frac{\left(x^2 - 1\right)\arctan(x^2)}{1 + x^4}\,\mathrm{d}x$$. – user149844 Sep 5 '14 at 2:09
• @user149844: The value to your integral in the comment is $\ln ( 2 ) \sqrt{2}\pi$ – Mhenni Benghorbal Sep 5 '14 at 2:45
• 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. – user149844 Sep 5 '14 at 4:54