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Question Number 193192 by 073 last updated on 07/Jun/23

$$\mathrm{solve}\mathrm{and}\mathrm{solution}$$$$\mathrm{\Omega }=\int \sqrt{{\sin }^{-1}\mathrm{x}}\mathrm{dx}=?$$

Answered by Frix last updated on 07/Jun/23

$$\int \sqrt{{\sin }^{-1}x}dx\stackrel{[t={\sin }^{-1}x]}{=}$$$$=\int \sqrt{t}\cos tdt\stackrel{\left[\mathrm{by}\mathrm{parts}\right]}{=}$$$$=\sqrt{t}\sin t-\frac{1}{2}\int \frac{\sin t}{\sqrt{t}}dt$$$$\frac{1}{2}\int \frac{\sin t}{\sqrt{t}}dt\stackrel{[u=\sqrt{\frac{2t}{\pi }}]}{=}$$$$=\sqrt{\frac{\pi }{2}}\int \sin \frac{\pi u^2}{2}du\stackrel{\left[\mathrm{Fresnel}\right]}{=}$$$$=\sqrt{\frac{\pi }{2}}\mathrm{S}\left(u\right)$$$$\Rightarrow$$$$\mathrm{\Omega }=x\sqrt{{\sin }^{-1}x}-\sqrt{\frac{\pi }{2}}\mathrm{S}\left(\sqrt{\frac{{2\sin }^{-1}x}{\pi }}\right)+C$$

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