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Question Number 189266 by cortano12 last updated on 14/Mar/23

$$\underset{0}{\stackrel{\pi /2}{\int }}\frac{\sqrt[3]{\tan \mathrm{x}}}{1+\sin 2\mathrm{x}}\mathrm{dx}=?$$

Answered by MJS_new last updated on 14/Mar/23

$$\int \frac{\sqrt[3]{\tan x}}{1+\sin 2x}dx=$$$$[t=\sqrt[3]{\tan x}\to dx={3\cos }^{2}x\sqrt[3]{{\tan }^{2}x}dt]$$$$=3\int \frac{t^3}{{(t^3+1)}^2}dt=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\right]$$$$=-\frac{t}{t^3+1}+\int \frac{dt}{t^3+1}=$$$$[\mathrm{decompose}\mathrm{etc}.]$$$$=-\frac{t}{t^3+1}+\frac{\sqrt{3}}{3}\arctan \frac{\sqrt{3}(2t-1)}{3}+\frac{1}{6}\ln \frac{{(t+1)}^2}{t^2-t+1}$$$$\mathrm{borders}\mathrm{for}t\mathrm{are}0⩽t<\mathrm{\infty }\Rightarrow$$$$\mathrm{answer}\mathrm{is}\frac{2\sqrt{3}\pi }{9}$$

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