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$$\int_{0}^{\pi/4}\ln(1+\tan x)\,dx \\$$ using $$\int_{0}^{a}f(x)\,dx = \int_{0}^{a}f(a-x)\,dx \\$$

I got up to $\int_{0}^{\pi/4}\ln(2)-\ln(1+\tan x)\,dx \\$ but I'm not sure how to integrate this

Thanks in advance

kjhg
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    If $$\int_{0}^{\frac{\pi}{4}} \ln(1 + \tan x) dx = \int_{0}^{\frac{\pi}{4}} \ln(2) - \ln(1+\tan x) dx$$ as you write, then rearranging gives $$2 \int_{0}^{\frac{\pi}{4}} \ln(1 + \tan x) dx = \int_{0}^{\frac{\pi}{4}} \ln(2) \implies \int_{0}^{\frac{\pi}{4}} \ln(1 + \tan x) dx = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \ln(2)$$ – Matthew Cassell Jul 17 '17 at 09:27
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    the result should be $$\frac{1}{8} \pi \log (2)$$ – Dr. Sonnhard Graubner Jul 17 '17 at 09:31
  • ahh I did not realise it was the same integral, thank you so much everyone for your help :) – kjhg Jul 17 '17 at 09:35

1 Answers1

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You are almost there:

$$\int_0^{\pi/4}\ln(1+\tan x)dx= \int_0^{\pi/4} \left(\ln(2) - \ln(1+\tan x)\right)dx$$

$$= \int_0^{\pi/4} \ln(2) dx - \int_0^{\pi/4} \ln(1+ \tan x) dx$$

$$\Rightarrow 2\int_0^{\pi/4}\ln(1+ \tan x)dx = \int_0^{\pi/4} \ln(2) dx$$ $$\Rightarrow \int_0^{\pi/4} \ln( 1 + \tan x)dx = \frac{\ln(2)}{2}\int_0^{\pi/4} dx = \boxed{\frac{\pi \ln(2)}{8}}$$