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Question Number 62093 by naka3546 last updated on 15/Jun/19

$$\int \frac{dx}{{\sin }^{3}x+{\cos }^{3}x}=p$$

Commented by MJS last updated on 15/Jun/19

$$\mathrm{Weierstrass}\mathrm{Substitution}$$$$\int \frac{dx}{{\sin }^{3}x+{\cos }^{3}x}=$$$$[t=\tan \frac{x}{2}\to dx=\frac{2dt}{1+t^2};\sin x=\frac{2t}{1+t^2};\cos x=\frac{1-t^2}{1+t^2}]$$$$=-2\int \frac{{(t^2+1)}^2}{(t^2-2t-1)(t^4+2t^3+2t^2-2t+1)}dt=$$$$=-2\int \frac{{(t^2+1)}^2}{(t-1-\sqrt{2})(t-1+\sqrt{2})(t^2+(1-\sqrt{3})t+2-\sqrt{3})(t^2+(1+\sqrt{3})t+2+\sqrt{3})}dt=$$$$=-\frac{\sqrt{2}}{3}\int \frac{dt}{t-1-\sqrt{2}}+\frac{\sqrt{2}}{3}\int \frac{dt}{t-1+\sqrt{2}}-\frac{1-\sqrt{3}}{3}\int \frac{dt}{t^2+(1-\sqrt{3})t+2-\sqrt{3}}-\frac{1+\sqrt{3}}{3}\int \frac{dt}{t^2+(1+\sqrt{3})t+2+\sqrt{3}}=$$$$=-\frac{\sqrt{2}}{3}\ln (t-1-\sqrt{2})+\frac{\sqrt{2}}{3}\ln (t-1+\sqrt{2})+\frac{2}{3}\arctan ((1+\sqrt{3})t-1)+\frac{2}{3}\arctan ((1-\sqrt{3})t-1)=$$$$=\frac{\sqrt{2}}{3}\ln \frac{t-1+\sqrt{2}}{t-1-\sqrt{2}}+\frac{2}{3}(\arctan ((1+\sqrt{3})t-1)+\arctan ((1-\sqrt{3})t-1))=$$$$=\frac{\sqrt{2}}{3}\ln \mid \frac{\tan \frac{x}{2}-1+\sqrt{2}}{\tan \frac{x}{2}-1-\sqrt{2}}\mid -\frac{2}{3}(\arctan (1-(1+\sqrt{3})\tan \frac{x}{2})+\arctan (1-(1-\sqrt{3})\tan \frac{x}{2}))+C$$

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