∫ (dx/( ((tan x))^(1/3) )) ?


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Question Number 124004 by john_santu last updated on 30/Nov/20

$\int \frac{d x}{\sqrt[3]{\tan x}} ?$
	Answered by liberty last updated on 30/Nov/20
	$T = ∫ (dx/( ((tan x))^(1/3) )) ; [ let u^3 =tan^2 x ∧ dx =((3u^2 )/(2u^(3/2) (u^3 +1)))du ] T = ∫ ((3u^2 )/(2u^(3/2) .u^(3/2) (u^3 +1))) du = (3/2)∫ (du/(u^3 +1)) T= (3/2)∫ (du/((u+1)(u^2 −u+1))) T=(1/2) {∫(du/(u+1)) − ∫ ((u−2)/(u^2 −u+1))du } T=(1/2)ℓn ∣u+1∣−(1/2)∫ ((2u−1−3)/(u^2 −u+1))du T=(1/2)ℓn ∣u+1∣−(1/2)ℓn∣u^2 −u+1∣ +(3/2)∫ (du/((u−(1/2))^2 +((1/2))^2 )) T=(1/2)ℓn ∣((u+1)/(u^2 −u+1))∣ +(3/2).2 arctan (((u−(1/2))/(1/2)))+c T=(1/2)ℓn ∣((u+1)/(u^2 −u+1))∣+3 arctan (2u−1)+c ∴ T = (1/2)ℓn ∣((((tan^2 x))^(1/3) +1)/( ((tan^4 x))^(1/3) −((tan^2 x))^(1/3) +1 ))∣ +3arctan (2((tan^2 x))^(1/3) −1 )+c$
	$T = \int \frac{d x}{\sqrt[3]{\tan x}}; [l e t u^{3} = \tan^{2} x \land d x = \frac{3 u^{2}}{2 u^{3 / 2} (u^{3} + 1)} d u]$ $T = \int \frac{3 u^{2}}{2 u^{3 / 2} . u^{3 / 2} (u^{3} + 1)} d u = \frac{3}{2} \int \frac{d u}{u^{3} + 1}$ $T = \frac{3}{2} \int \frac{d u}{(u + 1) (u^{2} - u + 1)}$ $T = \frac{1}{2} {\int \frac{d u}{u + 1} - \int \frac{u - 2}{u^{2} - u + 1} d u}$ $T = \frac{1}{2} ℓ n ∣ u + 1 ∣ - \frac{1}{2} \int \frac{2 u - 1 - 3}{u^{2} - u + 1} d u$ $T = \frac{1}{2} ℓ n ∣ u + 1 ∣ - \frac{1}{2} ℓ n ∣ u^{2} - u + 1 ∣ + \frac{3}{2} \int \frac{d u}{{(u - \frac{1}{2})}^{2} + {(\frac{1}{2})}^{2}}$ $T = \frac{1}{2} ℓ n ∣ \frac{u + 1}{u^{2} - u + 1} ∣ + \frac{3}{2} . 2 \arctan (\frac{u - \frac{1}{2}}{\frac{1}{2}}) + c$ $T = \frac{1}{2} ℓ n ∣ \frac{u + 1}{u^{2} - u + 1} ∣ + 3 \arctan (2 u - 1) + c$ $∴ T = \frac{1}{2} ℓ n ∣ \frac{\sqrt[3]{\tan^{2} x} + 1}{\sqrt[3]{\tan^{4} x} - \sqrt[3]{\tan^{2} x} + 1} ∣ + 3 \arctan (2 \sqrt[3]{\tan^{2} x} - 1) + c$
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