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Question Number 170025 by 0731619 last updated on 14/May/22

Answered by Mathspace last updated on 14/May/22

let (√(arcsinx))=t ⇒arsinx=t^2 ⇒x=sin(t^2 ) ⇒I=∫ ((2tcos(t^2 ))/t)dt =2∫ cos(t^2 )dt=2∫ Σ_(n=0) ^∞ (((−1)^n )/((2n)!))(t^2 )^(2n) dt =Σ_(n=0) ^∞ (((−1)^n )/((2n)!))∫ t^(4n) dt +C =Σ_(n=0) ^∞ (((−1)^n )/((4n+1)(2n)!))t^(4n+1) +C =Σ_(n=0) ^∞ (((−1)^n )/((4n+1)(2n)!))((√(arcsinx)))^(4n+1) +C

$${let}\:\sqrt{{arcsinx}}={t}\:\Rightarrow{arsinx}={t}^{\mathrm{2}} \\ $$$$\Rightarrow{x}={sin}\left({t}^{\mathrm{2}} \right)\:\Rightarrow{I}=\int\:\:\frac{\mathrm{2}{tcos}\left({t}^{\mathrm{2}} \right)}{{t}}{dt} \\ $$$$=\mathrm{2}\int\:{cos}\left({t}^{\mathrm{2}} \right){dt}=\mathrm{2}\int \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}\right)!}\left({t}^{\mathrm{2}} \right)^{\mathrm{2}{n}} {dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}\right)!}\int\:{t}^{\mathrm{4}{n}} {dt}\:+{C} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{4}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}\right)!}{t}^{\mathrm{4}{n}+\mathrm{1}} +{C} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{4}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}\right)!}\left(\sqrt{{arcsinx}}\right)^{\mathrm{4}{n}+\mathrm{1}} +{C} \\ $$

Commented by Mathspace last updated on 14/May/22

I=2Σ_(n=0) ^∞ (((−1)^n )/((4n+1)(2n)!))(arcsinx)^(2n+(1/2)) +C

$${I}=\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{4}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}\right)!}\left({arcsinx}\right)^{\mathrm{2}{n}+\frac{\mathrm{1}}{\mathrm{2}}} +{C} \\ $$

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