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Question Number 67359 by mhmd last updated on 26/Aug/19

∫siny/y  dy

$$\int{siny}/{y}\:\:{dy} \\ $$

Commented by mathmax by abdo last updated on 26/Aug/19

at form of serie  we have siny =Σ_(n=0) ^∞  (((−1)^n y^(2n+1) )/((2n+1)!)) ⇒  ((siny)/y) =Σ_(n=0) ^∞  (((−1)^n y^(2n) )/((2n+1)!)) ⇒ ∫ ((siny)/y)dy =Σ_(n=0) ^∞  (((−1)^n )/((2n+1)!(2n+1)))y^(2n+1)  +c

$${at}\:{form}\:{of}\:{serie}\:\:{we}\:{have}\:{siny}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {y}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\Rightarrow \\ $$$$\frac{{siny}}{{y}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {y}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\Rightarrow\:\int\:\frac{{siny}}{{y}}{dy}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!\left(\mathrm{2}{n}+\mathrm{1}\right)}{y}^{\mathrm{2}{n}+\mathrm{1}} \:+{c} \\ $$

Answered by Joel122 last updated on 26/Aug/19

Si(x) + C

$$\mathrm{Si}\left({x}\right)\:+\:{C} \\ $$

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