Question Number 6214 by sanusihammed last updated on 18/Jun/16
$$\int\sqrt{{tanx}}\:\:{dx}\: \\ $$
Commented by nburiburu last updated on 24/Jun/16
![by substitution t=(√(tan x))⇒t^2 = tan x 2t dt = sec^2 x dx = (1+tan^2 x) dx hence: dx =((2t)/(1+t^2 )) dt so I= ∫t . ((2t)/(1+t^2 )) dt =∫((2t^2 )/(1+t^2 )) dt = dividing = ∫ 2 − (2/(1+t^2 )) dt and it solves simply = 2(√(tan x))−2 atan[(√(tan x))]+c](https://www.tinkutara.com/question/Q6366.png)
$${by}\:\:{substitution} \\ $$$${t}=\sqrt{{tan}\:{x}}\Rightarrow{t}^{\mathrm{2}} =\:{tan}\:{x} \\ $$$$\mathrm{2}{t}\:{dt}\:=\:{sec}^{\mathrm{2}} {x}\:{dx}\:=\:\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)\:{dx} \\ $$$${hence}:\:{dx}\:=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt} \\ $$$${so} \\ $$$${I}=\:\int{t}\:.\:\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:=\int\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:=\: \\ $$$${dividing} \\ $$$$=\:\int\:\mathrm{2}\:−\:\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:{dt}\: \\ $$$${and}\:{it}\:{solves}\:{simply} \\ $$$$=\:\mathrm{2}\sqrt{{tan}\:{x}}−\mathrm{2}\:{atan}\left[\sqrt{{tan}\:{x}}\right]+{c} \\ $$