$$ \int_0^{\pi/2} \ln\left( \tan^2\left( \frac\pi4 + x \right) \right) \tan x\, dx $$ I tried to solve this question by substitution and let $u=\tan x$ And then using integration by parts or substitution But I want another method to solve it
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2If you succeeded in evaluating it, why do you need a second solution? – Chris Jun 25 '17 at 02:43
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@Chris OP may be looking for an easier method. – projectilemotion Jun 25 '17 at 16:11
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@projectilemotion How can any answerer write an "easier method", when the asker has refused to include his/her own work. – amWhy Jun 26 '17 at 12:25
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@amWhy I was thinking exactly the same, there is a lack of context in the question. It would have been a lot better if the derivation was shown explicitly, and would have made it a lot easier for us to provide an easier method. – projectilemotion Jun 26 '17 at 12:28
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I have evaluated it by series by letting u= tanx then break the interval from (0,1) and from (1,inf) then using a series to represent the integral withous using integration by parts – Mohammad Hussein Jun 26 '17 at 12:33
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@chris no need for your methods thanks alot – Mohammad Hussein Jun 26 '17 at 12:34
1 Answers
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** just a hint**
$$\int_0^\frac \pi 2=\int_0^\frac \pi 4+\int_\frac \pi 4^\frac \pi 2$$
$$\ln (\tan^2 (A ))=2\ln (|\tan (A)|) $$
$$\tan (x+\pi/4)=\frac {1+\tan (x)}{1-\tan (x)} $$
put $t=\tan (x) $
then by parts.
hamam_Abdallah
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