Tích phân từng phần: Đặt: $\begin{cases}u=x \\ dv=\frac{cos(x)}{sin^3(x)}dx \end{cases}\Rightarrow \begin{cases}du=dx \\ v=-\frac{1}{2sin^2(x)} \end{cases}$
$I=-\frac{x}{2sin^2(x)}|^{\pi/4}_{\pi/6}+\int\limits_{\pi/6}^{\pi/4}\frac{1}{2sin^2(x)}dx$
$=-\frac{\pi}{8sin^2(\pi/4)}+\frac{\pi}{12sin^2(\pi/6)}+\frac{1}{2}(-cot(x))|^{\pi/4}_{\pi/6}=\frac{\pi}{12}+\frac{\sqrt{3}-1}{2}$