Question Number 124004 by john_santu last updated on 30/Nov/20
$$\:\int\:\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}}}\:?\: \\ $$
Answered by liberty last updated on 30/Nov/20
![T = ∫ (dx/( ((tan x))^(1/3) )) ; [ let u^3 =tan^2 x ∧ dx =((3u^2 )/(2u^(3/2) (u^3 +1)))du ] T = ∫ ((3u^2 )/(2u^(3/2) .u^(3/2) (u^3 +1))) du = (3/2)∫ (du/(u^3 +1)) T= (3/2)∫ (du/((u+1)(u^2 −u+1))) T=(1/2) {∫(du/(u+1)) − ∫ ((u−2)/(u^2 −u+1))du } T=(1/2)ℓn ∣u+1∣−(1/2)∫ ((2u−1−3)/(u^2 −u+1))du T=(1/2)ℓn ∣u+1∣−(1/2)ℓn∣u^2 −u+1∣ +(3/2)∫ (du/((u−(1/2))^2 +((1/2))^2 )) T=(1/2)ℓn ∣((u+1)/(u^2 −u+1))∣ +(3/2).2 arctan (((u−(1/2))/(1/2)))+c T=(1/2)ℓn ∣((u+1)/(u^2 −u+1))∣+3 arctan (2u−1)+c ∴ T = (1/2)ℓn ∣((((tan^2 x))^(1/3) +1)/( ((tan^4 x))^(1/3) −((tan^2 x))^(1/3) +1 ))∣ +3arctan (2((tan^2 x))^(1/3) −1 )+c](https://www.tinkutara.com/question/Q124006.png)
$${T}\:=\:\int\:\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}}}\:;\:\left[\:{let}\:{u}^{\mathrm{3}} =\mathrm{tan}\:^{\mathrm{2}} {x}\:\wedge\:{dx}\:=\frac{\mathrm{3}{u}^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{3}/\mathrm{2}} \:\left({u}^{\mathrm{3}} +\mathrm{1}\right)}{du}\:\right] \\ $$$${T}\:=\:\int\:\frac{\mathrm{3}{u}^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{3}/\mathrm{2}} .{u}^{\mathrm{3}/\mathrm{2}} \left({u}^{\mathrm{3}} +\mathrm{1}\right)}\:{du}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\int\:\frac{{du}}{{u}^{\mathrm{3}} +\mathrm{1}} \\ $$$${T}=\:\frac{\mathrm{3}}{\mathrm{2}}\int\:\frac{{du}}{\left({u}+\mathrm{1}\right)\left({u}^{\mathrm{2}} −{u}+\mathrm{1}\right)} \\ $$$${T}=\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\int\frac{{du}}{{u}+\mathrm{1}}\:−\:\int\:\frac{{u}−\mathrm{2}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}{du}\:\right\} \\ $$$${T}=\frac{\mathrm{1}}{\mathrm{2}}\ell{n}\:\mid{u}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{2}{u}−\mathrm{1}−\mathrm{3}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}{du}\: \\ $$$${T}=\frac{\mathrm{1}}{\mathrm{2}}\ell{n}\:\mid{u}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}\ell{n}\mid{u}^{\mathrm{2}} −{u}+\mathrm{1}\mid\:+\frac{\mathrm{3}}{\mathrm{2}}\int\:\frac{{du}}{\left({u}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${T}=\frac{\mathrm{1}}{\mathrm{2}}\ell{n}\:\mid\frac{{u}+\mathrm{1}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}\mid\:+\frac{\mathrm{3}}{\mathrm{2}}.\mathrm{2}\:\mathrm{arctan}\:\left(\frac{{u}−\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}}}\right)+{c} \\ $$$${T}=\frac{\mathrm{1}}{\mathrm{2}}\ell{n}\:\mid\frac{{u}+\mathrm{1}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}\mid+\mathrm{3}\:\mathrm{arctan}\:\left(\mathrm{2}{u}−\mathrm{1}\right)+{c} \\ $$$$\therefore\:{T}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\ell{n}\:\mid\frac{\sqrt[{\mathrm{3}}]{\mathrm{tan}\:^{\mathrm{2}} {x}}\:\:+\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{tan}\:^{\mathrm{4}} {x}}\:−\sqrt[{\mathrm{3}}]{\mathrm{tan}\:^{\mathrm{2}} {x}}\:+\mathrm{1}\:}\mid\:+\mathrm{3arctan}\:\left(\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{tan}\:^{\mathrm{2}} {x}}\:−\mathrm{1}\:\right)+{c}\: \\ $$