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Question Number 81739 by TANMAY PANACEA last updated on 15/Feb/20

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

Commented by abdomathmax last updated on 16/Feb/20

$$I=\int \frac{dx}{{cos}^{3}x-{sin}^{3}x}=\int \frac{dx}{(cosx-sinx)({cos}^{2}x+cosxsinx+{sin}^{2}x)}$$$$=\int \frac{dx}{(cosx-sinx)(1+\frac{1}{2}sin\left(2x\right))}$$$$=\int \frac{2dx}{\sqrt{2}sin(x-\frac{\pi }{4})(2+sin\left(2x\right))}$$$${=}_{x-\frac{\pi }{4}=t}\sqrt{2}\int \frac{dt}{sin\left(t\right)(2+sin(2(t+\frac{\pi }{4})}$$$$=\sqrt{2}\int \frac{dt}{sint(2+cos\left(2t\right)}=\sqrt{2}\int \frac{dt}{sint(2+1-2{sin}^{2}t)}$$$$=\sqrt{2}\int \frac{dt}{sint(3-2{sin}^{2}t)}$$$$letdecomposeF\left(x\right)=\frac{1}{x(3-2x^2)}\Rightarrow$$$$F\left(x\right)=\frac{1}{x(\sqrt{3}-\sqrt{2}x)(\sqrt{3}+\sqrt{2}x)}$$$$=\frac{a}{x}+\frac{b}{\sqrt{3}-\sqrt{2}x}+\frac{c}{\sqrt{3}+\sqrt{2}x}$$$$a=\frac{1}{3}$$$$b=\frac{\sqrt{2}}{\sqrt{3}(\sqrt{3}+\sqrt{2}\times \frac{\sqrt{3}}{\sqrt{2}})}=\frac{\sqrt{2}}{6}$$$$c=-\frac{\sqrt{2}}{\sqrt{3}(\sqrt{3}-\sqrt{2}(-\frac{\sqrt{3}}{\sqrt{2}}))}=-\frac{\sqrt{2}}{6}\Rightarrow$$$$\Rightarrow I=\sqrt{2}\int F\left(sint\right)dt=\frac{1}{3}\int \frac{dt}{sint}+\frac{\sqrt{2}}{6}\int \frac{dt}{\sqrt{3}-\sqrt{2}sint}$$$$-\frac{\sqrt{2}}{6}\int \frac{dt}{\sqrt{3}+\sqrt{2}sint}integralsimpletofindby$$$$usingchangementtan\left(\frac{t}{2}\right)=u$$

Answered by MJS last updated on 15/Feb/20

$$\int \frac{dx}{{\cos }^{3}x-{\sin }^{3}x}=$$$$[t=\tan \frac{x}{2}\to dx=\frac{2dt}{t^2+1}]$$$$=-2\int \frac{{(t^2+1)}^2}{(t^2+2t-1)(t^2-(1+\sqrt{3})t+2+\sqrt{3})(t^2-(1-\sqrt{3})t+2-\sqrt{3})}dt=$$$$=-\frac{4}{3}\int \frac{dt}{t^2+2t-1}-\frac{1+\sqrt{3}}{3}\int \frac{dt}{t^2-(1+\sqrt{3})t+2+\sqrt{3}}-\frac{1-\sqrt{3}}{3}\int \frac{dt}{t^2-(1-\sqrt{3})t+2-\sqrt{3}}$$$$\mathrm{now}\mathrm{use}\mathrm{formula}$$

Commented by TANMAY PANACEA last updated on 15/Feb/20

$$excellentsir$$

Answered by TANMAY PANACEA last updated on 15/Feb/20

$$\int \frac{dx}{(cosx-sinx)(1+sinxcosx)}$$$$\int \frac{{(cosx-sinx)}^2+2sinxcosx+2-2}{(cosx-sinx)(1+cosxsinx)}dx$$$$\int \frac{cosx-sinx}{\frac{1}{2}[1+{(sinx+cosx)}^2]}dx+\int \frac{2dx}{cosx-sinx}-2I$$$$I=\frac{2}{3}\int \frac{d(sinx+cosx)}{1+{(sinx+cosx)}^2}+\frac{2}{3}\int \frac{\frac{1}{\sqrt{2}}}{sin(\frac{\pi }{4}-x)}$$$$I=\frac{2}{3}{tan}^{-1}(sinx+cosx)+\frac{\sqrt{2}}{3}\int cosec(\frac{\pi }{4}-x)dx$$$$I=\frac{2}{3}{tan}^{-1}(sinx+cosx)-\frac{\sqrt{2}}{3}lntan(\frac{\pi }{8}-\frac{x}{2})+c$$$$plscheckmistakeifany$$

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