Question

Evaluate $\int [\sqrt{\cot x}+\sqrt{\tan x}]dx$

Answer 1

$I=\int [\sqrt{cotx}+\sqrt{tanx}]dx$

$I=\int [\sqrt{\frac{cosx}{sinx}}+\sqrt{\frac{sinx}{cosx}}]dx$

$I=\int \frac{sinx+cosx}{\sqrt{sinx.cosx}}dx$

Let,

$sinx-cosx=u$ ; $u^2={(sinx-cosx)}^2={\sin }^{2}x+{\cos }^{2}x-2sinx.cosx$

$(sinx+cosx)dx=du$ $\Rightarrow u^2=1-2sinx.cosx$

$\Rightarrow sinx.cosx=\frac{1-u^2}{2}$

$\therefore$ $I=\int \frac{du}{\sqrt{\frac{1-u^2}{2}}}=\sqrt{2}\int \frac{du}{\sqrt{1-u^2}}=\sqrt{2}{\sin }^{-1}u+c$

$\therefore$ $I=\sqrt{2}{\sin }^{-1}(sinx-cosx)+c$

$=\sqrt{2}{\tan }^{-1}\left(\frac{sinx-cosx}{\sqrt{2sinx.cosx}}\right)+c$

$=\sqrt{2}{\tan }^{-1}\left(\frac{\frac{sinx}{cosx}-\frac{cosx}{cosx}}{\frac{1}{cosx}\sqrt{2sinx.cosx}}\right)+c$

$=\sqrt{2}{\tan }^{-1}\left(\frac{tanx-1}{\sqrt{\frac{2sinx.cosx}{{\cos }^{2}x}}}\right)+c$

$=\sqrt{2}{\tan }^{-1}\left(\frac{tanx-1}{\sqrt{2tanx}}\right)+c$