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Question Number 133786 by bobhans last updated on 24/Feb/21

$$\mathcal{A}={\int }_0^1{\sin }^{-1}\left(\frac{x^2+1}{\sqrt{2x^4+2}}\right)dx=?$$

Answered by john_santu last updated on 24/Feb/21

$$UsingthePythagoreantheorem$$$$\arcsin \left(\frac{x^2+1}{\sqrt{2x^4+2}}\right)=\arctan \left(\frac{1+x^2}{1-x^2}\right)$$$$\mathcal{A}={\int }_0^1\arctan \left(\frac{1+x^2}{1-x^2}\right)dx$$$$Usebyintegrationbyparts$$$$\mathcal{A}=x\arctan \left(\frac{1+x^2}{1-x^2}\right){\mid }_0^1-2{\int }_0^1\frac{x^2}{x^4+1}dx$$$$\mathcal{A}=\frac{\pi }{2}-[{\int }_0^1\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx+{\int }_0^1\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx]$$$$\mathcal{A}=\frac{\pi }{2}-\frac{1}{2\sqrt{2}}(\pi +\ln \left(3\right)-2\sqrt{2})$$$$\mathcal{A}\approx 1.0833$$

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