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Đặt: $I=\int\limits_{-1/2}^{1/2}\sin\left(\ln\frac{1-x}{1+x}\right)dx$
Đặt: $x=-t\Rightarrow dx=-dt$
Ta có:
$I=-\int\limits_{1/2}^{-1/2}\sin\left(\ln\frac{1+t}{1-t}\right)dt$
$=\int\limits_{-1/2}^{1/2}\sin\left(-\ln\frac{1-t}{1+t}\right)dt$
$=-\int\limits_{-1/2}^{1/2}\sin\left(\ln\frac{1-t}{1+t}\right)dt=-I$
Suy ra: $2I=0\Leftrightarrow I=0$
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