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

Answered by Mathspace last updated on 14/May/22

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

Commented by Mathspace last updated on 14/May/22

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

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