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Question Number 145456 by math55 last updated on 05/Jul/21

∫sin(x^2 +2)dx

$$\int\boldsymbol{{sin}}\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\right)\boldsymbol{{dx}} \\ $$

Answered by Olaf_Thorendsen last updated on 05/Jul/21

F(x) = ∫sin(x^2 +2)dx  F(x) = ∫(sin(x^2 )cos(2)+cos(x^2 )sin(2))dx  F(x) = cos(2)∫Σ_(n=0) ^∞ (−1)^n (((x^2 )^(2n+1) )/((2n+1)!)) dx  +sin(2)∫Σ_(n=0) ^∞ (−1)^n (((x^2 )^(2n) )/((2n)!))dx  F(x) = cos(2)Σ_(n=0) ^∞ (−1)^n (x^(4n+3) /((2n+1)!(4n+3)))   +sin(2)Σ_(n=0) ^∞ (−1)^n (x^(4n+1) /((2n)!(4n+1)))dx  F(x) = Σ_(n=0) ^∞ (−1)^n (x^(4n+1) /((2n)!)) (((cos(2)x^2 )/((2n+1)(4n+3)))+((sin(2))/(4n+1)))

$$\mathrm{F}\left({x}\right)\:=\:\int\mathrm{sin}\left({x}^{\mathrm{2}} +\mathrm{2}\right){dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int\left(\mathrm{sin}\left({x}^{\mathrm{2}} \right)\mathrm{cos}\left(\mathrm{2}\right)+\mathrm{cos}\left({x}^{\mathrm{2}} \right)\mathrm{sin}\left(\mathrm{2}\right)\right){dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{cos}\left(\mathrm{2}\right)\int\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\left({x}^{\mathrm{2}} \right)^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:{dx} \\ $$$$+\mathrm{sin}\left(\mathrm{2}\right)\int\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\left({x}^{\mathrm{2}} \right)^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{cos}\left(\mathrm{2}\right)\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{{x}^{\mathrm{4}{n}+\mathrm{3}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!\left(\mathrm{4}{n}+\mathrm{3}\right)}\: \\ $$$$+\mathrm{sin}\left(\mathrm{2}\right)\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{{x}^{\mathrm{4}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}\right)!\left(\mathrm{4}{n}+\mathrm{1}\right)}{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{{x}^{\mathrm{4}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}\right)!}\:\left(\frac{\mathrm{cos}\left(\mathrm{2}\right){x}^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{4}{n}+\mathrm{3}\right)}+\frac{\mathrm{sin}\left(\mathrm{2}\right)}{\mathrm{4}{n}+\mathrm{1}}\right) \\ $$

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