Question Number 84709 by M±th+et£s last updated on 15/Mar/20
$$\int\sqrt[{\mathrm{3}}]{{tan}\left({x}\right)}\:{dx} \\ $$
Answered by MJS last updated on 15/Mar/20
![∫((tan x))^(1/3) dx= [t=(tan x)^(2/3) → dx=(3/2)(sin x)^(1/3) (cos x)^(5/3) dt] =(3/2)∫(t/(t^3 +1))dt=(1/2)∫((t+1)/(t^2 −t+1))dt−(1/2)∫(dt/(t+1))= =(1/4)∫((2t−1)/(t^2 −t+1))dt+(3/4)∫(dt/(t^2 −t+1))−(1/2)∫(dt/(t+1))= =(1/4)ln (t^2 −t+1) +((√3)/2)arctan (((√3)/3)(2t−1)) −(1/2)ln (t+1) = =(1/4)ln ((t^2 −t+1)/((t+1)^2 )) +((√3)/2)arctan (((√3)/3)(2t−1)) now insert t=(tan x)^(2/3)](https://www.tinkutara.com/question/Q84719.png)
$$\int\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\left(\mathrm{tan}\:{x}\right)^{\mathrm{2}/\mathrm{3}} \:\rightarrow\:{dx}=\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{sin}\:{x}\right)^{\mathrm{1}/\mathrm{3}} \left(\mathrm{cos}\:{x}\right)^{\mathrm{5}/\mathrm{3}} {dt}\right] \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{t}}{{t}^{\mathrm{3}} +\mathrm{1}}{dt}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{t}+\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}+\mathrm{1}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{2}{t}−\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt}+\frac{\mathrm{3}}{\mathrm{4}}\int\frac{{dt}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}+\mathrm{1}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{arctan}\:\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{2}{t}−\mathrm{1}\right)\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({t}+\mathrm{1}\right)\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\frac{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{arctan}\:\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{2}{t}−\mathrm{1}\right)\right)\: \\ $$$$\mathrm{now}\:\mathrm{insert}\:{t}=\left(\mathrm{tan}\:{x}\right)^{\mathrm{2}/\mathrm{3}} \\ $$
Commented by M±th+et£s last updated on 15/Mar/20
$${thank}\:{you}\:{sir} \\ $$