solve-and-solution-sin-1-x-dx-

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

solve and solution

Ω = \int \sqrt{\sin^{- 1} x} dx = ?

Answered by Frix last updated on 07/Jun/23

\int \sqrt{\sin^{- 1} x} d x \overset{[t = \sin^{- 1} x]}{=}

= \int \sqrt{t} \cos t d t \overset{[by parts]}{=}

= \sqrt{t} \sin t - \frac{1}{2} \int \frac{\sin t}{\sqrt{t}} d t

\frac{1}{2} \int \frac{\sin t}{\sqrt{t}} d t \overset{[u = \sqrt{\frac{2 t}{π}}]}{=}

= \sqrt{\frac{π}{2}} \int \sin \frac{π u^{2}}{2} d u \overset{[Fresnel]}{=}

= \sqrt{\frac{π}{2}} S (u)

\Rightarrow

Ω = x \sqrt{\sin^{- 1} x} - \sqrt{\frac{π}{2}} S (\sqrt{\frac{{2 \sin}^{- 1} x}{π}}) + C

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