Question Number 113745 by Algoritm last updated on 15/Sep/20
Answered by MJS_new last updated on 15/Sep/20
![((2cos 2x −cos x)/(6−cos^2 x −4sin x))= =−((cos x)/(sin^2 x −4sin x +5))+((2(2cos^2 x −1))/(sin^2 x −4sin x +5)) −∫((cos x)/(sin^2 x −4sin x +5))dx= [t=2+sin x → dx=(dt/(cos x))] =−∫(dt/(t^2 +1))=−arctan t = =−arctan (2+sin x) 2∫((2cos^2 x −1)/(sin^2 x −4sin x +5))dx= [t=tan x +(√(1+tan^2 x)) → dx=((cos^2 x)/(1+sin x))dt] (x=arctan ((t^2 −1)/(2t))) =−2∫((t^4 −6t^2 +1)/((t^2 +1)(t^4 +4t^2 +5)))dt= =−2∫((t^4 −6t^2 +1)/((t^2 +1)(t^2 −(√(−4+2(√5))) t+(√5))(t^2 +(√(−4+2(√5))) t+(√5))))dt now decompose and use the formulas for ∫((ax+b)/(x^2 +cx+d))dx. I think you should start to work for yourself now...](Q113783.png)
$$\frac{\mathrm{2cos}\:\mathrm{2}{x}\:−\mathrm{cos}\:{x}}{\mathrm{6}−\mathrm{cos}^{\mathrm{2}} \:{x}\:−\mathrm{4sin}\:{x}}= \\ $$$$=−\frac{\mathrm{cos}\:{x}}{\mathrm{sin}^{\mathrm{2}} \:{x}\:−\mathrm{4sin}\:{x}\:+\mathrm{5}}+\frac{\mathrm{2}\left(\mathrm{2cos}^{\mathrm{2}} \:{x}\:−\mathrm{1}\right)}{\mathrm{sin}^{\mathrm{2}} \:{x}\:−\mathrm{4sin}\:{x}\:+\mathrm{5}} \\ $$$$ \\ $$$$−\int\frac{\mathrm{cos}\:{x}}{\mathrm{sin}^{\mathrm{2}} \:{x}\:−\mathrm{4sin}\:{x}\:+\mathrm{5}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{2}+\mathrm{sin}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{cos}\:{x}}\right] \\ $$$$=−\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}=−\mathrm{arctan}\:{t}\:= \\ $$$$=−\mathrm{arctan}\:\left(\mathrm{2}+\mathrm{sin}\:{x}\right) \\ $$$$ \\ $$$$\mathrm{2}\int\frac{\mathrm{2cos}^{\mathrm{2}} \:{x}\:−\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:{x}\:−\mathrm{4sin}\:{x}\:+\mathrm{5}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}\:\rightarrow\:{dx}=\frac{\mathrm{cos}^{\mathrm{2}} \:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}{dt}\right] \\ $$$$\:\:\:\:\:\:\:\left({x}=\mathrm{arctan}\:\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{t}}\right) \\ $$$$=−\mathrm{2}\int\frac{{t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} +\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{4}} +\mathrm{4}{t}^{\mathrm{2}} +\mathrm{5}\right)}{dt}= \\ $$$$=−\mathrm{2}\int\frac{{t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} +\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} −\sqrt{−\mathrm{4}+\mathrm{2}\sqrt{\mathrm{5}}}\:{t}+\sqrt{\mathrm{5}}\right)\left({t}^{\mathrm{2}} +\sqrt{−\mathrm{4}+\mathrm{2}\sqrt{\mathrm{5}}}\:{t}+\sqrt{\mathrm{5}}\right)}{dt} \\ $$$$\mathrm{now}\:\mathrm{decompose}\:\mathrm{and}\:\mathrm{use}\:\mathrm{the}\:\mathrm{formulas}\:\mathrm{for} \\ $$$$\int\frac{{ax}+{b}}{{x}^{\mathrm{2}} +{cx}+{d}}{dx}.\:\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{should}\:\mathrm{start}\:\mathrm{to} \\ $$$$\mathrm{work}\:\mathrm{for}\:\mathrm{yourself}\:\mathrm{now}... \\ $$