Question

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

Answer 1

$I={\int }_{\left(\frac{\pi }{4}\right)}^{\left(\frac{\pi }{3}\right)}(tanx+cotx{)}^{2}dx={\int }_{\left(\frac{\pi }{4}\right)}^{\left(\frac{\pi }{3}\right)}(ta{n}^{2}x+co{t}^{2}x+2tanxcotx)dx={\int }_{\left(\frac{\pi }{4}\right)}^{\left(\frac{\pi }{3}\right)}(ta{n}^{2}x+co{t}^{2}x+2)dx={\int }_{\left(\frac{\pi }{4}\right)}^{\left(\frac{\pi }{3}\right)}\left[\right(se{c}^{2}x-1+cose{c}^2-1)+2]dx={[tanx-cotx]}_{\pi /6}^{\pi /3}=\left(tan\right(\frac{\pi }{3})-cot(\frac{\pi }{3}\left)\right)-\left(tan\right(\frac{\pi }{6})-cot(\frac{\pi }{6}\left)\right)=(\sqrt{3}-(\frac{1}{\sqrt{3}}\left)\right)-\left(\right(\frac{1}{\sqrt{3}})-\sqrt{3})=2(\sqrt{3}-(\frac{1}{\sqrt{3}}\left)\right)=2\left(\right(\frac{3-1}{\sqrt{3}}\left)\right)=(\frac{4}{\sqrt{3}})$