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Question Number 200685 by Spillover last updated on 21/Nov/23

$$$$$${\int }_0^{\frac{\pi }{4}}\ln (1+\tan x)dx$$$$$$

Answered by som(math1967) last updated on 22/Nov/23

$$I={\int }_0^{\frac{\pi }{4}}ln[1+tan(\frac{\pi }{4}-x)]dx$$$$I={\int }_0^{\frac{\pi }{4}}ln[1+\frac{1-tanx}{1+tanx}]dx$$$$I={\int }_0^{\frac{\pi }{4}}ln\left(\frac{2}{1+tanx}\right)dx$$$$I={\int }_0^{\frac{\pi }{4}}ln2dx-{\int }_0^{\frac{\pi }{4}}ln(1+tanx)dx$$$$I={\int }_0^{\frac{\pi }{4}}ln2dx-I$$$$2I=ln2{\left[x\right]}_0^{\frac{\pi }{4}}$$$$I=\frac{\pi }{8}ln2$$

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