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Question Number 84709 by M±th+et£s last updated on 15/Mar/20

$$\int \sqrt[3]{tan\left(x\right)}dx$$

Answered by MJS last updated on 15/Mar/20

$$\int \sqrt[3]{\tan x}dx=$$$$[t={\left(\tan x\right)}^{2/3}\to dx=\frac{3}{2}{\left(\sin x\right)}^{1/3}{\left(\cos x\right)}^{5/3}dt]$$$$=\frac{3}{2}\int \frac{t}{t^3+1}dt=\frac{1}{2}\int \frac{t+1}{t^2-t+1}dt-\frac{1}{2}\int \frac{dt}{t+1}=$$$$=\frac{1}{4}\int \frac{2t-1}{t^2-t+1}dt+\frac{3}{4}\int \frac{dt}{t^2-t+1}-\frac{1}{2}\int \frac{dt}{t+1}=$$$$=\frac{1}{4}\ln (t^2-t+1)+\frac{\sqrt{3}}{2}\arctan \left(\frac{\sqrt{3}}{3}(2t-1)\right)-\frac{1}{2}\ln (t+1)=$$$$=\frac{1}{4}\ln \frac{t^2-t+1}{{(t+1)}^2}+\frac{\sqrt{3}}{2}\arctan \left(\frac{\sqrt{3}}{3}(2t-1)\right)$$$$\mathrm{now}\mathrm{insert}t={\left(\tan x\right)}^{2/3}$$

Commented by M±th+et£s last updated on 15/Mar/20

$$thankyousir$$

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