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Đặt $t=a+b-x \implies dt=-dx$
Khi $x=a \Rightarrow t=b, x=b \Rightarrow t=a$ .
Như vậy $\int\limits_{a}^{b}f(a+b-x)dx=-\int\limits_{b}^{a}f(t)dt=\int\limits_{a}^{b}f(t)dt=\int\limits_{a}^{b}f(x)dx$ (đpcm).
Áp dụng ta có
$I=\int\limits_{0}^{\frac{\pi}{4} }\ln (1+\tan x)dx$
$I=\int\limits_{0}^{\frac{\pi}{4} }\ln (1+\tan \left (\frac{\pi}{4} -x\right ))dx$
$I=\int\limits_{0}^{\frac{\pi}{4} }\ln (1+\frac{1-\tan x}{1+\tan x})dx$
$I=\int\limits_{0}^{\frac{\pi}{4} }\ln (\frac{2}{1+\tan x})dx$
$I=\int\limits_{0}^{\frac{\pi}{4} }\ln 2dx-\int\limits_{0}^{\frac{\pi}{4} }\ln (1+\tan x)dx=\int\limits_{0}^{\frac{\pi}{4} }\ln 2dx-I$
$\implies 2I=\int\limits_{0}^{\frac{\pi}{4} }\ln 2dx \implies I=\frac{\pi}{8}\ln 2$
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