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Question Number 113745 by Algoritm last updated on 15/Sep/20

Answered by MJS_new last updated on 15/Sep/20

$$\frac{2\cos 2x-\cos x}{6-{\cos }^{2}x-4\sin x}=$$$$=-\frac{\cos x}{{\sin }^{2}x-4\sin x+5}+\frac{2({2\cos }^{2}x-1)}{{\sin }^{2}x-4\sin x+5}$$$$$$$$-\int \frac{\cos x}{{\sin }^{2}x-4\sin x+5}dx=$$$$[t=2+\sin x\to dx=\frac{dt}{\cos x}]$$$$=-\int \frac{dt}{t^2+1}=-\arctan t=$$$$=-\arctan (2+\sin x)$$$$$$$$2\int \frac{{2\cos }^{2}x-1}{{\sin }^{2}x-4\sin x+5}dx=$$$$[t=\tan x+\sqrt{1+{\tan }^{2}x}\to dx=\frac{{\cos }^{2}x}{1+\sin x}dt]$$$$(x=\arctan \frac{t^2-1}{2t})$$$$=-2\int \frac{t^4-6t^2+1}{(t^2+1)(t^4+4t^2+5)}dt=$$$$=-2\int \frac{t^4-6t^2+1}{(t^2+1)(t^2-\sqrt{-4+2\sqrt{5}}t+\sqrt{5})(t^2+\sqrt{-4+2\sqrt{5}}t+\sqrt{5})}dt$$$$\mathrm{now}\mathrm{decompose}\mathrm{and}\mathrm{use}\mathrm{the}\mathrm{formulas}\mathrm{for}$$$$\int \frac{ax+b}{x^2+cx+d}dx.\mathrm{I}\mathrm{think}\mathrm{you}\mathrm{should}\mathrm{start}\mathrm{to}$$$$\mathrm{work}\mathrm{for}\mathrm{yourself}\mathrm{now}...$$

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