Mathematics

Find 
$$\displaystyle \int_0^{1/4 \pi} ln(1 + \tan x )dx$$.


SOLUTION
$$I = \displaystyle \int_0^{\dfrac{\pi}{4}} \ln (1 + \tan x) dx$$
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$$I = \displaystyle\int_0^{\dfrac{\pi}{4}} \ln \left(1 + \tan \left(\dfrac{\pi}{4} - 4\right)\right) dx$$
$$= \displaystyle\int_0^{\dfrac{\pi}{4}} \ln \left(1 + \dfrac{1 - \tan x}{1 + \tan x} \right) dx$$
$$= \displaystyle\int_0^{\dfrac{\pi}{4}} \ln \left(\dfrac{1 + \tan x + 1 - \tan x}{1 + \tan x } \right) dx$$
$$= \displaystyle\int_0^{\dfrac{\pi}{4}} \ln \left(\dfrac{2}{1 + \tan x} \right) dx$$
$$2 I = \displaystyle\int_0^{\dfrac{\pi}{4}} \ln \left(\dfrac{2}{1 + \tan x} \right) + \ln (1 + \tan x ) dx$$
$$2 I = \displaystyle\int_0^{\dfrac{\pi}{4}} \ln (2) dx$$
$$2I = \ln (2) \left(\dfrac{\pi}{4} - 0\right)$$
$$I = \dfrac{\pi}{8} \ln (2)$$
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Subjective Medium Published on 17th 09, 2020
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