Question Number 193192 by 073 last updated on 07/Jun/23
$$\mathrm{solve}\:\mathrm{and}\:\mathrm{solution} \\ $$$$\Omega=\int\sqrt{\mathrm{sin}^{−\mathrm{1}} \mathrm{x}}\mathrm{dx}=? \\ $$
Answered by Frix last updated on 07/Jun/23
![∫(√(sin^(−1) x)) dx =^([t=sin^(−1) x]) =∫(√t) cos t dt =^([by parts]) =(√t) sin t −(1/2)∫((sin t)/( (√t)))dt (1/2)∫((sin t)/( (√t)))dt =^([u=(√((2t)/π))]) =(√(π/2))∫sin ((πu^2 )/2) du =^([Fresnel]) =(√(π/2))S (u) ⇒ Ω=x(√(sin^(−1) x)) −(√(π/2)) S ((√((2sin^(−1) x)/π))) +C](https://www.tinkutara.com/question/Q193217.png)
$$\int\sqrt{\mathrm{sin}^{−\mathrm{1}} \:{x}}\:{dx}\:\overset{\left[{t}=\mathrm{sin}^{−\mathrm{1}} \:{x}\right]} {=} \\ $$$$=\int\sqrt{{t}}\:\mathrm{cos}\:{t}\:{dt}\:\overset{\left[\mathrm{by}\:\mathrm{parts}\right]} {=} \\ $$$$=\sqrt{{t}}\:\mathrm{sin}\:{t}\:−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{sin}\:{t}}{\:\sqrt{{t}}}{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{sin}\:{t}}{\:\sqrt{{t}}}{dt}\:\overset{\left[{u}=\sqrt{\frac{\mathrm{2}{t}}{\pi}}\right]} {=} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\sqrt{\frac{\pi}{\mathrm{2}}}\int\mathrm{sin}\:\frac{\pi{u}^{\mathrm{2}} }{\mathrm{2}}\:{du}\:\overset{\left[\mathrm{Fresnel}\right]} {=} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\sqrt{\frac{\pi}{\mathrm{2}}}\mathrm{S}\:\left({u}\right) \\ $$$$\Rightarrow \\ $$$$\Omega={x}\sqrt{\mathrm{sin}^{−\mathrm{1}} \:{x}}\:−\sqrt{\frac{\pi}{\mathrm{2}}}\:\mathrm{S}\:\left(\sqrt{\frac{\mathrm{2sin}^{−\mathrm{1}} \:{x}}{\pi}}\right)\:+{C} \\ $$