Answer
$$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{sin\theta\,cot\theta}{sec\theta}d\theta=\frac{\pi}{12}$$
Work Step by Step
$$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{sin\theta\,cot\theta}{sec\theta}d\theta=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{sin\theta\frac{cos\theta}{sin\theta}}{\frac{1}{cos\theta}}d\theta=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}cos^{2}\theta d\theta$$
$$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}
\frac{1+cos2\theta}{2}d\theta=\left [\frac{\theta}{2}+\frac{sin2\theta}{4}\right ]_{\frac{\pi}{6}}^{\frac{\pi}{3}}$$
$$=\frac{\pi}{12}$$
