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Question Number 117910 by bemath last updated on 14/Oct/20

$$\int {\sin }^{-1}\left(\sqrt{\mathrm{x}}\right)\mathrm{dx}=?$$

Commented by bemath last updated on 14/Oct/20

Answered by Lordose last updated on 14/Oct/20

$$$$$$\int {\sin }^{-1}\left(\sqrt{\mathrm{x}}\right)\mathrm{dx}=?$$$$\mathrm{u}=\sqrt{\mathrm{x}}\Rightarrow \mathrm{dx}=2\mathrm{udu}$$$$2\int {\mathrm{usin}}^{-1}\left(\mathrm{u}\right)\mathrm{du}$$$$\mathrm{IBP}$$$$2(\frac{{\mathrm{u}}^2}{2}{\sin }^{-1}\left(\mathrm{u}\right)-\frac{1}{2}\int \frac{{\mathrm{u}}^2}{\sqrt{1-{\mathrm{u}}^2}}\mathrm{du})$$$${\mathrm{u}}^2{\sin }^{-1}\left(\mathrm{u}\right)-\int \frac{{\sin }^2\theta \cos \theta \mathrm{d}\theta }{\cos \theta }\{\mathrm{u}=\sin \theta \}$$$${\mathrm{u}}^2{\sin }^{-1}\left(\mathrm{u}\right)-\int {\sin }^2\theta \mathrm{d}\theta$$$${\mathrm{u}}^2{\sin }^{-1}\left(\mathrm{u}\right)-(\frac{\theta }{2}-\frac{\sin \theta \cos \theta }{2})+\mathrm{C}$$$${\mathrm{u}}^2{\sin }^{-1}\left(\mathrm{u}\right)-\frac{{\sin }^{-1}\left(\mathrm{u}\right)}{2}+\frac{\mathrm{u}\sqrt{1-{\mathrm{u}}^2}}{2}+\mathrm{C}$$$$\frac{1}{2}{\sin }^{-1}\left(\sqrt{\mathrm{x}}\right)(2\mathrm{x}-1)+\frac{\sqrt{\mathrm{x}-{\mathrm{x}}^2}}{2}+\mathrm{C}$$

Answered by Dwaipayan Shikari last updated on 14/Oct/20

$$\int 2{usin}^{-1}udux=u^2\Rightarrow 1=2u\frac{du}{dx}$$$$=u^2{sin}^{-1}u-\int \frac{u^2}{\sqrt{1-u^2}}du$$$$=u^2{sin}^{-1}u+\int \sqrt{1-u^2}-\frac{1}{\sqrt{1-u^2}}du$$$$=u^2{sin}^{-1}u+\int cos\theta \sqrt{1-{sin}^2\theta }d\theta -{sin}^{-1}uu=sin\theta$$$$=(x-1){sin}^{-1}\left(\sqrt{x}\right)+\frac{1}{2}\theta +\frac{1}{2}\int cos2\theta d\theta$$$$=(x-1){sin}^{-1}\left(\sqrt{x}\right)+\frac{1}{2}{sin}^{-1}\left(\sqrt{x}\right)+\frac{1}{4}sin\left(2{sin}^{-1}\sqrt{x}\right)+C$$$$=\frac{1}{2}(2x-1){sin}^{-1}\left(\sqrt{x}\right)+\frac{1}{4}sin\left(2{sin}^{-1}\sqrt{x}\right)+C$$$$★sin\left(2\theta \right)=2sin\theta cos\theta =2\sqrt{x}\sqrt{1-x}=2\sqrt{x-x^2}$$$$$$

Answered by 1549442205PVT last updated on 14/Oct/20

$$\mathrm{Find}\mathrm{F}=\int {\sin }^{-1}\left(\sqrt{\mathrm{x}}\right)\mathrm{dx}=$$$$\mathrm{Put}\sqrt{\mathrm{x}}=\mathrm{t}\Rightarrow \mathrm{dt}=\frac{\mathrm{dx}}{2\sqrt{\mathrm{x}}}.\Rightarrow \mathrm{dx}=2\mathrm{tdt}$$$$\mathrm{F}=\int {\sin }^{-1}\left(\sqrt{\mathrm{x}}\right)\mathrm{dx}=\int {2\mathrm{tsin}}^{-1}\left(\mathrm{t}\right)\mathrm{dt}$$$$\text{Missing left or extra right}$$$$={\mathrm{t}}^2{\sin }^{-1}\left(\mathrm{t}\right)+\int \frac{1-{\mathrm{t}}^2}{\sqrt{1-{\mathrm{t}}^2}}\mathrm{dt}-\int \frac{\mathrm{dt}}{\sqrt{1-{\mathrm{t}}^2}}$$$$={\mathrm{t}}^2{\sin }^{-1}\left(\mathrm{t}\right)-{\sin }^{-1}\left(\mathrm{t}\right)+\int \sqrt{1-{\mathrm{t}}^2}\mathrm{dt}\left(1\right)$$$$\mathrm{Put}\mathrm{t}=\mathrm{sinu}\Rightarrow \mathrm{dt}=\mathrm{cosudu},\sqrt{1-{\mathrm{t}}^2}=\mathrm{cosu}$$$$\Rightarrow \int \sqrt{1-{\mathrm{t}}^2}\mathrm{dt}=\int {\cos }^2\mathrm{udu}=\frac{1}{2}\int (1+\cos 2\mathrm{u})\mathrm{du}$$$$=\frac{\mathrm{u}}{2}+\frac{1}{4}\sin 2\mathrm{u}=\frac{1}{2}{\sin }^{-1}\mathrm{t}+\frac{1}{2}\mathrm{sinucosu}$$$$=\frac{1}{2}{\sin }^{-1}\left(\mathrm{t}\right)+\frac{1}{2}\mathrm{t}\sqrt{1-{\mathrm{t}}^2}\left(2\right)$$$$\mathrm{From}\left(1\right)\left(2\right)\mathrm{we}\mathrm{get}:$$$$\mathrm{F}={\mathrm{t}}^2{\sin }^{-1}\left(\mathrm{t}\right)-\frac{1}{2}{\sin }^{-1}\left(\mathrm{t}\right)+\frac{1}{2}\mathrm{t}\sqrt{1-{\mathrm{t}}^2}$$$$=(\mathrm{x}-\frac{1}{2}){\sin }^{-1}\left(\sqrt{\mathrm{x}}\right)+\frac{1}{2}\sqrt{\mathrm{x}(1-\mathrm{x})}+\mathrm{C}$$

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