Question

$$\underset{\pi /6}{\overset{\pi /3}{\int }}\frac{1}{1+\sqrt{\tan x}}dx$$

Answer 1

$$LetI={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{1}{1+\sqrt{\tan x}}dx.................(1)$$
$$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{1}{1+\sqrt{\tan (\frac{\pi }{3}+\frac{\pi }{6}-x)}}dx$$
$$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{1}{1+\sqrt{cotx}}dx....................(2)$$
$$\text{Adding (1) and (2)}$$
$$2I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}(\frac{1}{1+\sqrt{\tan x}}+\frac{1}{1+\sqrt{cotx}})dx$$
$$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\left(\frac{1+\sqrt{cotx}+1+\sqrt{\tan x}}{1+\sqrt{cotx}+\sqrt{\tan x}+\sqrt{\tan xcotx}}\right)dx$$
$$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{2+\sqrt{cotx}+\sqrt{\tan x}}{2+\sqrt{cotx}+\sqrt{\tan x}}dx$$
$$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}dx={\left[x\right]}_{\frac{\pi }{6}}^{\frac{\pi }{3}}$$
$$=\frac{\pi }{3}-\frac{\pi }{6}$$
$$\therefore 2I=\frac{\pi }{6}$$
$$HenceI=\frac{\pi }{12}$$