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Question Number 144826 by mathdanisur last updated on 29/Jun/21

Answered by mindispower last updated on 29/Jun/21

$${sin}^{-}\left(a\right)+{sin}^{-}\left(b\right)=u$$$$a,b\in [0,1]$$$$sin\left(u\right)=a\sqrt{1-b^2}+b\sqrt{1-a^2}$$$$u={sin}^{-}(a\sqrt{1-b^2}+b\sqrt{1-a^2})$$$$a=\frac{x}{\sqrt{x^2+y^2}},b=\frac{y}{\sqrt{x^2+y^2}}$$$${sin}^{-}\left(\frac{x}{\sqrt{x^2+y^2}}\right)+{sin}^{-}\left(\frac{y}{\sqrt{x^2+y^2}}\right)$$$$={sin}^{-}(\frac{x}{\sqrt{x^2+y^2}}.\sqrt{1-\frac{y^2}{x^2+y^2}}+\frac{y}{\sqrt{x^2+y^2}}.\sqrt{1-\frac{x^2}{x^2+y^2}})$$$$={sin}^{-}\left(\frac{x^2+y^2}{x^2+y^2}\right)=\frac{\pi }{2}$$$$\iff {\int }_a^{a+1}{\int }_a^{a+1}\frac{\pi }{2}dxdy=\frac{\pi }{2}$$$$$$$$$$

Commented by mathdanisur last updated on 29/Jun/21

$$alotperfectSirthanks$$

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