Question Number 71695 by peter frank last updated on 18/Oct/19
Answered by MJS last updated on 19/Oct/19
![∫((√(tan x))/(1+(√(tan x))))dx= [t=(√(tan x)) → dx=2(√(tan x))cos^2 x dt=((2t)/(t^4 +1))dt] =2∫(t^2 /((t+1)(t^4 +1)))dt= =2∫(t^2 /((t+1)(t^2 −(√2)t+1)(t^2 +(√2)t+1)))dt= =∫(dt/(t+1))−((1−(√2))/2)∫((t+1)/(t^2 −(√2)t+1))dt−((1+(√2))/2)∫((t+1)/(t^2 +(√2)t+1))dt= =ln (t+1) − −((1−(√2))/4)ln (t^2 −(√2)t+1) +(1/2)arctan ((√2)t−1) −y −((1+(√2))/4)ln (t^2 +(√2)t+1) −(1/2)arctan ((√2)t+1) and now put t=(√(tan x))](Q71703.png)
$$\int\frac{\sqrt{\mathrm{tan}\:{x}}}{\mathrm{1}+\sqrt{\mathrm{tan}\:{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{tan}\:{x}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{\mathrm{tan}\:{x}}\mathrm{cos}^{\mathrm{2}} \:{x}\:{dt}=\frac{\mathrm{2}{t}}{{t}^{\mathrm{4}} +\mathrm{1}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{4}} +\mathrm{1}\right)}{dt}= \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)}{dt}= \\ $$$$=\int\frac{{dt}}{{t}+\mathrm{1}}−\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{t}+\mathrm{1}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}}{dt}−\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{t}+\mathrm{1}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}}{dt}= \\ $$$$=\mathrm{ln}\:\left({t}+\mathrm{1}\right)\:− \\ $$$$\:\:\:\:\:−\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{t}−\mathrm{1}\right)\:−{y} \\ $$$$\:\:\:\:\:−\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\left({t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{t}+\mathrm{1}\right) \\ $$$$\mathrm{and}\:\mathrm{now}\:\mathrm{put}\:{t}=\sqrt{\mathrm{tan}\:{x}} \\ $$
Commented by peter frank last updated on 19/Oct/19
$${thank}\:{you} \\ $$