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Question Number 202125 by Calculusboy last updated on 21/Dec/23

$$\int \frac{\mathit{s}\mathit{i}\mathit{n}\left(3\mathit{x}\right)}{1+{\mathit{s}\mathit{i}\mathit{n}}^3\mathit{x}}\mathit{d}\mathit{x}$$

Answered by Frix last updated on 21/Dec/23

$$\mathrm{Let}s=\sin x$$$$\frac{\sin 3x}{1+{\sin }^{3}x}=\frac{s(4s^2-3)}{(s+1)(s^2-s+1)}=$$$$=-4+\frac{1}{3(s+1)}-\frac{s-11}{3(s^2-s+1)}$$$$\int \frac{\sin 3x}{1+{\sin }^{3}x}dx=$$$$=-4\int dx+\frac{1}{3}\int \frac{dx}{1+\sin x}+\frac{1}{3}\int \frac{11-\sin x}{1-\sin x+{\sin }^{2}x}dx$$$$\mathrm{These}\mathrm{are}\mathrm{easy}:$$$$-4\int dx=-4x$$$$\frac{1}{3}\int \frac{dx}{1+\sin x}=-\frac{1-\sin x}{3\cos x}$$$$\mathrm{This}\mathrm{is}\mathrm{unpleasant}:$$$$\frac{1}{3}\int \frac{11-\sin x}{1-\sin x+{\sin }^{2}x}dx$$$$\mathrm{I}\mathrm{tried}t=\tan \frac{x}{2}\mathrm{but}...$$$$\mathrm{Instead}\mathrm{I}\mathrm{solved}\mathrm{it}\mathrm{with}$$$$t=\tan x+\frac{1}{\cos x}\iff x={\tan }^{-1}\frac{t^2-1}{2t}\to dx=\frac{2dt}{t^2+1}$$$$\mathrm{which}\mathrm{leads}\mathrm{to}$$$$\frac{4}{3}\int \frac{5t^2+6}{t^4+3}dt=\frac{4}{3}\int \frac{5t^2+6}{(t^2-\sqrt[4]{12}t+\sqrt{3})(t^2+\sqrt[4]{12}t+\sqrt{3})}dt$$$$\mathrm{Now}\mathrm{decompose}...$$

Commented by Calculusboy last updated on 22/Dec/23

$$\mathit{n}\mathit{i}\mathit{c}\mathit{e}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}$$

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