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Question Number 92949 by Ar Brandon last updated on 09/May/20
∫(1/x)sin((1/x)) dx
$$\int\frac{\mathrm{1}}{\mathrm{x}}\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:\mathrm{dx} \\ $$
Commented by mathmax by abdo last updated on 10/May/20
I =∫ (1/x)sin((1/x))dx changement (1/x)=t give I =∫t sint (−(dt/t^2 )) =−∫ ((sint)/t)dt we have sint =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!))t^(2n+1) ⇒((sint)/t) =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!)) t^(2n) ⇒∫ ((sint)/t)dt =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)×(2n+1)!))t^(2n+1) +c =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)(2n+1)!))×(1/x^(2n+1) ) +c =1−(1/(3×3!))×(1/x^3 ) +(1/(5×5!))×(1/x^5 )−.... +c
$${I}\:=\int\:\frac{\mathrm{1}}{{x}}{sin}\left(\frac{\mathrm{1}}{{x}}\right){dx}\:\:{changement}\:\frac{\mathrm{1}}{{x}}={t}\:{give} \\ $$$${I}\:=\int{t}\:{sint}\:\left(−\frac{{dt}}{{t}^{\mathrm{2}} }\right)\:=−\int\:\:\frac{{sint}}{{t}}{dt}\:\:{we}\:{have}\:{sint}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{t}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\Rightarrow\frac{{sint}}{{t}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:{t}^{\mathrm{2}{n}} \:\Rightarrow\int\:\frac{{sint}}{{t}}{dt}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)×\left(\mathrm{2}{n}+\mathrm{1}\right)!}{t}^{\mathrm{2}{n}+\mathrm{1}} \:+{c} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)!}×\frac{\mathrm{1}}{{x}^{\mathrm{2}{n}+\mathrm{1}} }\:+{c} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}×\mathrm{3}!}×\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:+\frac{\mathrm{1}}{\mathrm{5}×\mathrm{5}!}×\frac{\mathrm{1}}{{x}^{\mathrm{5}} }−….\:+{c} \\ $$
Commented by mathmax by abdo last updated on 10/May/20
I =(1/x)−(1/(3.3! x^3 ))+(1/(5.5! x^5 )) +....+c
$${I}\:=\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{3}.\mathrm{3}!\:{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{5}.\mathrm{5}!\:{x}^{\mathrm{5}} }\:+….+{c} \\ $$
Answered by peter frank last updated on 10/May/20
let u=(1/x) du=−(1/x^2 )dx dx=−x^2 du ∫usinu .−x^2 du −∫(1/u)sin udu sin u=u−(u^3 /(3!))+(u^5 /(5!))−(u^7 /(7!)) ((sin u)/u)=1−(u^2 /(3!))+(u^4 /(5!))−(u^6 /(7!)) −∫(1−(u^2 /(3!))+(u^4 /(5!))−(u^6 /(7!)))du −(u−(u^3 /(3.3!))+(u^5 /(5.5!))−(u^7 /(7.7!))) u=(1/x) −(u−(u^3 /(3.3!))+(u^5 /(5.5!))−(u^7 /(7.7!)))
$${let}\:{u}=\frac{\mathrm{1}}{{x}} \\ $$$${du}=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{dx} \\ $$$${dx}=−{x}^{\mathrm{2}} {du} \\ $$$$\int{u}\mathrm{sin}{u}\:.−{x}^{\mathrm{2}} {du} \\ $$$$−\int\frac{\mathrm{1}}{{u}}\mathrm{sin}\:{udu} \\ $$$$\mathrm{sin}\:{u}={u}−\frac{{u}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{u}^{\mathrm{5}} }{\mathrm{5}!}−\frac{{u}^{\mathrm{7}} }{\mathrm{7}!} \\ $$$$\frac{\mathrm{sin}\:{u}}{{u}}=\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{u}^{\mathrm{4}} }{\mathrm{5}!}−\frac{{u}^{\mathrm{6}} }{\mathrm{7}!} \\ $$$$−\int\left(\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{u}^{\mathrm{4}} }{\mathrm{5}!}−\frac{{u}^{\mathrm{6}} }{\mathrm{7}!}\right){du} \\ $$$$−\left({u}−\frac{{u}^{\mathrm{3}} }{\mathrm{3}.\mathrm{3}!}+\frac{{u}^{\mathrm{5}} }{\mathrm{5}.\mathrm{5}!}−\frac{{u}^{\mathrm{7}} }{\mathrm{7}.\mathrm{7}!}\right) \\ $$$$\:{u}=\frac{\mathrm{1}}{{x}} \\ $$$$−\left({u}−\frac{{u}^{\mathrm{3}} }{\mathrm{3}.\mathrm{3}!}+\frac{{u}^{\mathrm{5}} }{\mathrm{5}.\mathrm{5}!}−\frac{{u}^{\mathrm{7}} }{\mathrm{7}.\mathrm{7}!}\right) \\ $$$$ \\ $$

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