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b) Đặt $t=-x\Rightarrow dt=-dx$
$I= \int\limits_{-\pi/4}^{\pi/4} \frac{\sin^{6}x + \cos^{6}x}{6^{x}+1}dx= \int\limits_{\pi/4}^{-\pi/4} \frac{\sin^{6}(-t) + \cos^{6}(-t)}{6^{-t}+1}(-dt)= \int\limits_{-\pi/4}^{\pi/4} \frac{(\sin^{6}t + \cos^{6}t)6^t}{1+6^{t}}dt$
$\implies I= \int\limits_{-\pi/4}^{\pi/4} \frac{(\sin^{6}x + \cos^{6}x)6^x}{1+6^{x}}dx$
Suy ra
$2I=I+I = \int\limits_{-\pi/4}^{\pi/4} \frac{\sin^{6}x + \cos^{6}x}{6^{x}+1}dx+ \int\limits_{-\pi/4}^{\pi/4} \frac{(\sin^{6}x + \cos^{6}x)6^x}{1+6^{x}}dx= \int\limits_{-\pi/4}^{\pi/4}(\sin^{6}x + \cos^{6}x)dx$
$= \int\limits_{-\pi/4}^{\pi/4}(\dfrac{5}{8}+\dfrac{3}{8}\cos 4x)dx=\left[ {\dfrac{5x}{8}+\dfrac{3}{32}\sin 4x} \right]_{-\pi/4}^{\pi/4}=\dfrac{5\pi}{16}$
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