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Question Number 51990 by maxmathsup by imad last updated on 01/Jan/19

$$calculate{\int }_{\frac{1}{2}}^{1}xarctan\left(\sqrt{1-x^2}\right)dx$$

Answered by peter frank last updated on 01/Jan/19

$$bypart$$$$v=\frac{x^2}{2}$$$$u={\tan }^{-1}\left(\sqrt{1-x^2}\right)$$$$\frac{du}{dx}=-\frac{2x}{2-x^2}$$$$\frac{x^2}{2}{\tan }^{-1}\left(\sqrt{1-x^2}\right)+\int \frac{x^3}{2-x^2}dx$$$$\frac{x^2}{2}{\tan }^{-1}\left(\sqrt{1-x^2}\right)+\int \frac{2x-x^3+2x}{2-x^2}$$$$\frac{x^2}{2}{\tan }^{-1}\left(\sqrt{1-x^2}\right)+ln(2-x^2)+C$$$$....$$

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