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Evaluate-sinx-x-4-x-2-1-dx-I-need-full-detailed-explanation-thank-you-in-advance-

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Question Number 204804 by Mastermind last updated on 27/Feb/24
Evaluate ∫((sinx)/(x^4 +x^2 +1))dx I need full detailed explanation, thank you in advance.
$$\mathrm{Evaluate}\:\int\frac{\mathrm{sinx}}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{need}\:\mathrm{full}\:\mathrm{detailed}\:\mathrm{explanation},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{in} \\ $$$$\mathrm{advance}. \\ $$
Commented by TonyCWX08 last updated on 28/Feb/24
I would use the Contour Integration but thete is no interval...
$${I}\:{would}\:{use}\:{the}\:{Contour}\:{Integration}\:{but}\:{thete}\:{is}\:{no}\:{interval}… \\ $$
Commented by Mastermind last updated on 28/Feb/24
Kindly evaluate... or what do you think we can do ?
$$\mathrm{Kindly}\:\mathrm{evaluate}…\:\mathrm{or}\:\mathrm{what}\:\mathrm{do}\:\mathrm{you}\:\mathrm{think}\:\mathrm{we} \\ $$$$\mathrm{can}\:\mathrm{do}\:? \\ $$
Commented by TonyCWX08 last updated on 28/Feb/24
Contour Integral is used in Complex World Usually used for the integral with interval (−∞,∞) But in this case, it can′t be used.
$$ \\ $$$${Contour}\:{Integral}\:{is}\:{used}\:{in}\:{Complex}\:{World} \\ $$$${Usually}\:{used}\:{for}\:{the}\:{integral}\:{with}\:{interval}\:\left(−\infty,\infty\right) \\ $$$${But}\:{in}\:{this}\:{case},\:{it}\:{can}'{t}\:{be}\:{used}. \\ $$
Answered by Frix last updated on 28/Feb/24
((sin x)/(x^4 +x^2 +1))=((sin x)/((x^2 −x+1)(x^2 +x+1)))= =(((1−x)sin x)/(x^2 −x+1))+(((1+x)sin x)/(x^2 +x+1))= =((sin x)/(x^2 −x+1))−((xsin x)/(x^2 −x+1))+((sin x)/(x^2 +x+1))+((xsin x)/(x^2 +x+1))= =((4sin x)/((2x−1−(√3)i)(2x−1+(√3)i)))− −((4xsin x)/((2x−1−(√3)i)(2x−1+(√3)i)))+ +((4sin x)/((2x+1−(√3)i)(2x+1+(√3)i)))+ +((4xsin x)/((2x+1−(√3)i)(2x+1+(√3)i)))= =((2isin x)/( (√3)(2x−1+(√3)i)))−((2isin x)/( (√3)(2x−1−(√3)i)))+ +(((1−(√3)i)sin x)/( (√3)(2x−1−(√3)i)))−((((√3)+i)sin x)/( (√3)(2x−1+(√3)i)))− −((2isin x)/( (√3)(2x+1−(√3)i)))+((2isin x)/( (√3)(2x+1+(√3)i)))+ +((((√3)+i)sin x)/( (√3)(2x+1−(√3)i)))+((((√3)−i)sin x)/( (√3)(2x+1+(√3)i))) (Please check for sign errors) So essentially we need to solve integrals of the shape a∫((sin x)/(bx+c))dx=(a/b)(cos (c/b) Si (x+(c/b)) −sin (c/b) Ci (x+(c/b)))+C with Si (Ci) being the Integral (Co)Sine ∫((sin x)/x)dx=Si x +C ∫((cos x)/x)dx=Ci x +C Somebody finish this please.
$$\frac{\mathrm{sin}\:{x}}{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{sin}\:{x}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}= \\ $$$$=\frac{\left(\mathrm{1}−{x}\right)\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}+\frac{\left(\mathrm{1}+{x}\right)\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}= \\ $$$$=\frac{\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}−\frac{{x}\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}+\frac{\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\frac{{x}\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}= \\ $$$$=\frac{\mathrm{4sin}\:{x}}{\left(\mathrm{2}{x}−\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{i}\right)\left(\mathrm{2}{x}−\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}\right)}− \\ $$$$−\frac{\mathrm{4}{x}\mathrm{sin}\:{x}}{\left(\mathrm{2}{x}−\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{i}\right)\left(\mathrm{2}{x}−\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}\right)}+ \\ $$$$+\frac{\mathrm{4sin}\:{x}}{\left(\mathrm{2}{x}+\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{i}\right)\left(\mathrm{2}{x}+\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}\right)}+ \\ $$$$+\frac{\mathrm{4}{x}\mathrm{sin}\:{x}}{\left(\mathrm{2}{x}+\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{i}\right)\left(\mathrm{2}{x}+\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}\right)}= \\ $$$$=\frac{\mathrm{2isin}\:{x}}{\:\sqrt{\mathrm{3}}\left(\mathrm{2}{x}−\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}\right)}−\frac{\mathrm{2isin}\:{x}}{\:\sqrt{\mathrm{3}}\left(\mathrm{2}{x}−\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{i}\right)}+ \\ $$$$+\frac{\left(\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{i}\right)\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{3}}\left(\mathrm{2}{x}−\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{i}\right)}−\frac{\left(\sqrt{\mathrm{3}}+\mathrm{i}\right)\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{3}}\left(\mathrm{2}{x}−\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}\right)}− \\ $$$$−\frac{\mathrm{2isin}\:{x}}{\:\sqrt{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{i}\right)}+\frac{\mathrm{2isin}\:{x}}{\:\sqrt{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}\right)}+ \\ $$$$+\frac{\left(\sqrt{\mathrm{3}}+\mathrm{i}\right)\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{i}\right)}+\frac{\left(\sqrt{\mathrm{3}}−\mathrm{i}\right)\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}\right)} \\ $$$$\left(\mathrm{Please}\:\mathrm{check}\:\mathrm{for}\:\mathrm{sign}\:\mathrm{errors}\right) \\ $$$$\mathrm{So}\:\mathrm{essentially}\:\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{integrals} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{shape} \\ $$$${a}\int\frac{\mathrm{sin}\:{x}}{{bx}+{c}}{dx}=\frac{{a}}{{b}}\left(\mathrm{cos}\:\frac{{c}}{{b}}\:\mathrm{Si}\:\left({x}+\frac{{c}}{{b}}\right)\:−\mathrm{sin}\:\frac{{c}}{{b}}\:\mathrm{Ci}\:\left({x}+\frac{{c}}{{b}}\right)\right)+{C} \\ $$$$\mathrm{with}\:\mathrm{Si}\:\left(\mathrm{Ci}\right)\:\mathrm{being}\:\mathrm{the}\:\mathrm{Integral}\:\left(\mathrm{Co}\right)\mathrm{Sine} \\ $$$$\int\frac{\mathrm{sin}\:{x}}{{x}}{dx}=\mathrm{Si}\:{x}\:+{C}\:\:\:\:\:\int\frac{\mathrm{cos}\:{x}}{{x}}{dx}=\mathrm{Ci}\:{x}\:+{C} \\ $$$$\mathrm{Somebody}\:\mathrm{finish}\:\mathrm{this}\:\mathrm{please}. \\ $$

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