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Question Number 49638 by maxmathsup by imad last updated on 08/Dec/18

$$leta>2andf\left(a\right)={\int }_{-\frac{1}{a}}^{\frac{1}{a}}\frac{x^{2}dx}{\sqrt{1+x^2}+\sqrt{1-x^2}}$$ $$1)calculatef\left(a\right)intermsofa$$ $$2)calculate{f}^{{}^{\prime }}\left(a\right).$$

Answered by Smail last updated on 08/Dec/18

$$f\left(a\right)={\stackrel{1/a}{\int }}_{-1/a}\frac{x^2(\sqrt{1+x^2}-\sqrt{1-x^2})}{(1+x^2)-(1-x^2)}dx$$ $$={\int }_{-1/a}^{1/a}\frac{x^2(\sqrt{1+x^2}-\sqrt{1-x^2})}{2x^2}dx$$ $$=\frac{1}{2}{\int }_{-1/a}^{1/a}(\sqrt{1+x^2}-\sqrt{1-x^2})dx$$ $$\frac{1}{2}{\int }_{-1/a}^{1/a}\sqrt{1+x^2}dx-\frac{1}{2}{\int }_{-1/a}^{1/a}\sqrt{1-x^2}dx$$ $$x=sinh\left(t\right)andx=sin\left(u\right)$$ $$f\left(a\right)=\frac{1}{2}{\int }_{{sinh}^{-1}(-1/a)}^{{sinh}^{-1}\left(1/a\right)}{cosh}^2\left(t\right)dt-\frac{1}{2}{\int }_{{sin}^{-1}(-1/a)}^{{sin}^{-1}\left(1/a\right)}{cos}^2\left(u\right)du$$ $$=\frac{1}{4}{[\frac{sinh\left(2t\right)}{2}+t]}_{{sinh}^{-1}(-1/a)}^{{sinh}^{-1}\left(1/a\right)}-\frac{1}{4}{[\frac{cos\left(2u\right)}{2}+u]}_{{sin}^{-1}(-1/a)}^{{sin}^{-1}\left(1/a\right)}$$ $$=\frac{1}{2}(\frac{1}{a}\sqrt{1+{\left(\frac{1}{a}\right)}^2}+{sinh}^{-1}\left(1/a\right))-\frac{1}{2}(\frac{1}{a}\sqrt{1-\frac{1}{a^2}}+{sin}^{-1}\left(1/a\right))$$ $$f\left(a\right)=\frac{1}{2}(\frac{\sqrt{a^2+1}}{a^2}+{sinh}^{-1}\left(\frac{1}{a}\right)-\frac{\sqrt{a^2-1}}{a^2}-{sin}^{-1}\left(1/a\right))$$ $$$$

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