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$=\int\limits_{0}^{\frac{\Pi }{4}}dx-\int\limits_{0}^{\frac{\Pi }{4}}tan^8xdx$
$=x|_0^\frac{\Pi }{4} -\int\limits_{0}^{\frac{\Pi }{4}}\frac{sin^8x}{cos^8x}dx$ $= \frac{\Pi }{4}-A.$
$A=\int\limits_{0}^{\frac{\Pi }{4}}\frac{sin^8x.cosx}{cos^9x}dx$
Đặt $t = cos^9x=>dt=-9.sinx.cos^8xdx$
$=> A = -\frac{1}{9}\int\limits_{0}^{\frac{\Pi }{4}}tdt=-\frac{1}{9}\ln |t||_0^\frac{\Pi }{4}$
Bạn tự lắp cận vào nhé
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