Question Number 61667 by aliesam last updated on 06/Jun/19
$$\int\sqrt{{tan}\left({x}\right)}\:{dx}\: \\ $$
Answered by MJS last updated on 06/Jun/19
![∫(√(tan x))dx= [t=(√(tan x)) → dx=2cos^2 x (√(tan x))dt] =2∫(t^2 /(t^4 +1))dt=2∫(t^2 /((t^2 −(√2)t+1)(t^2 +(√2)t+1)))dt= =(√2)∫((t/(t^2 −(√2)t+1))−(t/(t^2 +(√2)t+1)))dt= =∫((((√2)t−1)/(t^2 −(√2)t+1))+(1/(t^2 −(√2)t+1))−(((√2)t+1)/(t^2 +(√2)t+1))+(1/(t^2 −(√2)t+1)))dt= =((√2)/2)ln ((t^2 −(√2)t+1)/(t^2 +(√2)t+1)) +(√2)(arctan ((√2)t−1) +arctan ((√2)t+1))+C](https://www.tinkutara.com/question/Q61676.png)
$$\int\sqrt{\mathrm{tan}\:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{tan}\:{x}}\:\rightarrow\:{dx}=\mathrm{2cos}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{tan}\:{x}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} +\mathrm{1}}{dt}=\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)}{dt}= \\ $$$$=\sqrt{\mathrm{2}}\int\left(\frac{{t}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}}−\frac{{t}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}}\right){dt}= \\ $$$$=\int\left(\frac{\sqrt{\mathrm{2}}{t}−\mathrm{1}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}}−\frac{\sqrt{\mathrm{2}}{t}+\mathrm{1}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}}\right){dt}= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{ln}\:\frac{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}}\:+\sqrt{\mathrm{2}}\left(\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{t}−\mathrm{1}\right)\:+\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)\right)+{C} \\ $$
Commented by aliesam last updated on 06/Jun/19
$${well}\:{done}\:{sir} \\ $$