Question

$${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}{(\tan x+\cot x)}^{2}dx$$

Answer 1

$${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}{(\tan x+\cot x)}^{2}dx$$
$$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}({\tan }^{2}x+{\cot }^{2}x+2\tan x\cot x)dx$$
$$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}({\sec }^{2}x-1+{cosec}^{2}x-1+2)dx$$
$$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\sec }^{2}xdx+{\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}{cosec}^{2}xdx$$

$$={\tan x|}_{\frac{\pi }{6}}^{\frac{\pi }{3}}+{(-\cot x)|}_{\frac{\pi }{6}}^{\frac{\pi }{3}}$$
$$=(\tan \frac{\pi }{3}-\tan \frac{\pi }{6})-(\cot \frac{\pi }{3}-\cot \frac{\pi }{6})$$
$$=(\sqrt{3}-\frac{1}{\sqrt{3}})-(\frac{1}{\sqrt{3}}-\sqrt{3})$$
$$=2\sqrt{3}-\frac{2}{\sqrt{3}}$$
$$=\frac{4}{\sqrt{3}}$$