Đặt $u=x-\frac{\pi}{3}\Rightarrow du=dx , u\in[-\frac{\pi}{3},0]$$\int\limits_{0}^{\frac{\pi}{3}}\frac{sinx}{cos^3(x-\frac{\pi}{3})}dx=\int\limits_{-\frac{\pi}{3}}^{0}\frac{sin(u+\frac{\pi}{3})}{cos^3u}du=\int\limits_{-\frac{\pi}{3}}^{0}\frac{\frac{1}{2}sinu+\frac{\sqrt3}{2}cosu}{cos^3u}du$
$=\frac{1}{2}\int\limits_{-\frac{\pi}{3}}^{0}\frac{sinu.du}{cos^3u}+\frac{\sqrt3}{2}\int\limits_{-\frac{\pi}{3}}^{0}\frac{du}{cos^2u}$
$=\frac{1}{2}\int\limits_{-\frac{\pi}{3}}^{0}-\frac{d(cosu)}{cos^3u}+\frac{\sqrt3}{2}\int\limits_{-\frac{\pi}{3}}^{0}tanu$
$=\frac{1}{4}\int\limits_{-\frac{\pi}{3}}^{0}\frac{1}{cos^2u}+\frac{\sqrt3}{2}(tan0-tan-\frac{\pi}{3})$
$=\frac{1}{4}(\frac{1}{cos^20}-\frac{1}{cos^2-\frac{\pi}{3}})+\frac{\sqrt3}{2}\sqrt3$
$=\frac{1}{4}(1-4)+\frac{3}{2}$
$=-\frac{3}{4}+\frac{3}{2}$
$=\frac{3}{4}$