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Question Number 122751 by bemath last updated on 19/Nov/20

$$\underset{0}{\stackrel{1/\sqrt{2}}{\int }}\frac{x{\sin }^{-1}\left(x^2\right)}{\sqrt{1-x^4}}dx?$$

Answered by liberty last updated on 19/Nov/20

$$Let{\sin }^{-1}\left(x^2\right)=uthendu=\frac{2xdx}{\sqrt{1-x^4}}$$$$andintegralbecomes\frac{1}{2}\int udu=\frac{1}{4}{u}^2+c$$$$thus\underset{0}{\stackrel{1/\sqrt{2}}{\int }}\frac{x{\sin }^{-1}\left(x^2\right)}{\sqrt{1-x^4}}dx=\frac{1}{4}\left({\sin }^{-1}\left(x^2\right)\right){\mid }_0^{1/\sqrt{2}}$$$$=\frac{{\pi }^2}{144}.$$

Answered by Dwaipayan Shikari last updated on 19/Nov/20

$${\int }_0^{\frac{1}{\sqrt{2}}}\frac{{xsin}^{-1}\left(x^2\right)}{\sqrt{1-x^4}}dx{x}^2=t\Rightarrow 2x=\frac{dt}{dx}$$$$\frac{1}{2}{\int }_0^{\frac{1}{2}}\frac{{sin}^{-1}\left(t\right)}{\sqrt{1-t^2}}dt=\frac{1}{4}{\left[{\left({sin}^{-1}t\right)}^2\right]}_0^{\frac{1}{2}}=\frac{1}{4}.\frac{{\pi }^2}{36}=\frac{{\pi }^2}{144}$$

Answered by Bird last updated on 19/Nov/20

$$I={\int }_0^{\frac{1}{\sqrt{2}}}\frac{xarcsin\left(x^2\right)}{\sqrt{1-x^4}}dxwedothe$$$$changement{x}^2=t\Rightarrow 2xdx=dt\Rightarrow$$$$I={\int }_0^{\frac{1}{2}}\frac{arcsin\left(t\right)}{\sqrt{1-t^2}}\frac{dt}{2}$$$$\Rightarrow 2I={\left[{arcsin}^{2}t\right]}_0^{\frac{1}{2}}-{\int }_0^{\frac{1}{2}}\frac{arcsint}{\sqrt{1-t^2}}dt$$$$3I={\left(\frac{\pi }{6}\right)}^2=\frac{{\pi }^2}{36}\Rightarrow I=\frac{{\pi }^2}{3.36}$$$$=\frac{{\pi }^2}{108}$$

Commented by liberty last updated on 19/Nov/20

$$no.theanswer\frac{{\pi }^2}{144}$$

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