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Question Number 87046 by Tajaddin last updated on 02/Apr/20

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

Answered by MJS last updated on 02/Apr/20

$$\int \frac{dx}{{\sin }^{3}x+{\cos }^{3}x}=$$$$[t=\tan \frac{x}{2}\to dx=\frac{2dt}{t^2+1}]$$$$=-2\int \frac{t^4+2t^2+1}{t^6-3t^4-8t^3+3t^2-1}dt=$$$$=-2\int \frac{{(t^2+1)}^2}{(t^2-2t-1)(t^4+2t^3+2t^2-2t+1)}dt=$$$$=-2\int \frac{{(t^2+1)}^2}{(t^2-1-\sqrt{2})(t-1+\sqrt{2})(t^2+(1-\sqrt{3})t+2-\sqrt{3})(t^2+(1+\sqrt{3})t+2+\sqrt{3})}dt=$$$$=-\frac{\sqrt{2}}{3}\int \frac{dt}{t-1-\sqrt{2}}+\frac{\sqrt{2}}{3}\int \frac{dt}{t-1+\sqrt{2}}-$$$$-\frac{1-\sqrt{3}}{3}\int \frac{dt}{t^2+(1-\sqrt{3})t+2-\sqrt{3}}-$$$$-\frac{1+\sqrt{3}}{3}\int \frac{dt}{t^2+(1+\sqrt{3})t+2+\sqrt{3}}=$$$$\left[\mathrm{now}\mathrm{using}\mathrm{formula}\right]$$$$=\frac{\sqrt{2}}{3}\ln \frac{t-1+\sqrt{2}}{t-1-\sqrt{2}}+\frac{2}{3}(\arctan ((1+\sqrt{3})t-1)+\arctan ((1-\sqrt{3})t-1)$$$$\mathrm{now}\mathrm{insert}t=\tan \frac{x}{2}$$

Commented by Ar Brandon last updated on 02/Apr/20

$$Niceone,butSir...howdidyouperform$$$$thefactorisationsoeasily?Howdidyoudothat?$$

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