How to integrate $\int {\sqrt {\tan x} } dx$ ~~~~~[NCERT Exemp. Q. 43,Page 166 ]

Let $I = \int {\sqrt {\tan x} } dx$


Let’s put $\tan x = {t^2} \Rightarrow {\sec ^2}xdx = 2tdt$


therefore,$I = \int t \cdot \frac{{2t}}{{{{\sec }^2}x}}dt = 2\int {\frac{{{t^2}}}{{1 + {t^4}}}} dt$


$ = \int {\frac{{\left( {{t^2} + 1} \right) + \left( {{t^2} – 1} \right)}}{{\left( {1 + {t^4}} \right)}}} dt$

$ = \int {\frac{{{t^2} + 1}}{{1 + {t^4}}}} dt + \int {\frac{{{t^2} – 1}}{{1 + {t^4}}}} dt$

$ = \int {\frac{{1 + \frac{1}{{{t^2}}}}}{{{t^2} + \frac{1}{{{t^2}}}}}} dt + \int {\frac{{1 – \frac{1}{{{t^2}}}}}{{{t^2} + \frac{1}{{{t^2}}}}}} dt$

$ = \int {\frac{{1 – \left( { – \frac{1}{{{t^2}}}} \right)dt}}{{{{\left( {t – \frac{1}{t}} \right)}^2} + 2}}} + \int {\frac{{1 + \left( { – \frac{1}{{{t^2}}}} \right)}}{{{{\left( {t + \frac{1}{t}} \right)}^2} – 2}}} dt$


Let’s put $u = t – \frac{1}{t} \Rightarrow du = \left( {1 + \frac{1}{{{t^2}}}} \right)dt$

and $v = t + \frac{1}{t} \Rightarrow dv = \left( {1 – \frac{1}{{{t^2}}}} \right)dt$


therefore,$I = \int {\frac{{du}}{{{u^2} + {{(\sqrt 2 )}^2}}}} + \int {\frac{{dv}}{{{v^2} – {{(\sqrt 2 )}^2}}}} $


$ = \frac{1}{{\sqrt 2 }}{\tan ^{ – 1}}\frac{u}{{\sqrt 2 }} + \frac{1}{{2\sqrt 2 }}\log \left| {\frac{{v – \sqrt 2 }}{{v + \sqrt 2 }}} \right| + C$


$ = \frac{1}{{\sqrt 2 }}{\tan ^{ – 1}}\left( {\frac{{\tan x – 1}}{{\sqrt {2\tan x} }}} \right) + \frac{1}{{2\sqrt 2 }}\log \left| {\frac{{\tan x – \sqrt {2\tan x} + 1}}{{\tan x + \sqrt {2\tan x} + 1}}} \right| + C$

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