Question Number 49638 by maxmathsup by imad last updated on 08/Dec/18
$${let}\:{a}>\mathrm{2}\:{and}\:{f}\left({a}\right)\:=\int_{−\frac{\mathrm{1}}{{a}}} ^{\frac{\mathrm{1}}{{a}}} \:\:\:\frac{{x}^{\mathrm{2}} {dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({a}\right)\:{interms}\:{of}\:{a} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{f}^{'} \left({a}\right)\:. \\ $$
Answered by Smail last updated on 08/Dec/18
![f(a)=∫_(−1/a) ^(1/a) ((x^2 ((√(1+x^2 ))−(√(1−x^2 ))))/((1+x^2 )−(1−x^2 )))dx =∫_(−1/a) ^(1/a) ((x^2 ((√(1+x^2 ))−(√(1−x^2 ))))/(2x^2 ))dx =(1/2)∫_(−1/a) ^(1/a) ((√(1+x^2 ))−(√(1−x^2 )))dx (1/2)∫_(−1/a) ^(1/a) (√(1+x^2 ))dx−(1/2)∫_(−1/a) ^(1/a) (√(1−x^2 ))dx x=sinh(t) and x=sin(u) f(a)=(1/2)∫_(sinh^(−1) (−1/a)) ^(sinh^(−1) (1/a)) cosh^2 (t)dt−(1/2)∫_(sin^(−1) (−1/a)) ^(sin^(−1) (1/a)) cos^2 (u)du =(1/4)[((sinh(2t))/2)+t]_(sinh^(−1) (−1/a)) ^(sinh^(−1) (1/a)) −(1/4)[((cos(2u))/2)+u]_(sin^(−1) (−1/a)) ^(sin^(−1) (1/a)) =(1/2)((1/a)(√(1+((1/a))^2 ))+sinh^(−1) (1/a))−(1/2)((1/a)(√(1−(1/a^2 )))+sin^(−1) (1/a)) f(a)=(1/2)(((√(a^2 +1))/a^2 )+sinh^(−1) ((1/a))−((√(a^2 −1))/a^2 )−sin^(−1) (1/a))](https://www.tinkutara.com/question/Q49659.png)
$${f}\left({a}\right)=\overset{\mathrm{1}/{a}} {\int}_{−\mathrm{1}/{a}} \frac{{x}^{\mathrm{2}} \left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{dx} \\ $$$$=\int_{−\mathrm{1}/{a}} ^{\mathrm{1}/{a}} \frac{{x}^{\mathrm{2}} \left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{\mathrm{2}{x}^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{1}/{a}} ^{\mathrm{1}/{a}} \left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right){dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{1}/{a}} ^{\mathrm{1}/{a}} \sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{1}/{a}} ^{\mathrm{1}/{a}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$${x}={sinh}\left({t}\right)\:\:\:{and}\:\:{x}={sin}\left({u}\right) \\ $$$${f}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int_{{sinh}^{−\mathrm{1}} \left(−\mathrm{1}/{a}\right)} ^{{sinh}^{−\mathrm{1}} \left(\mathrm{1}/{a}\right)} {cosh}^{\mathrm{2}} \left({t}\right){dt}−\frac{\mathrm{1}}{\mathrm{2}}\int_{{sin}^{−\mathrm{1}} \left(−\mathrm{1}/{a}\right)} ^{{sin}^{−\mathrm{1}} \left(\mathrm{1}/{a}\right)} {cos}^{\mathrm{2}} \left({u}\right){du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{{sinh}\left(\mathrm{2}{t}\right)}{\mathrm{2}}+{t}\right]_{{sinh}^{−\mathrm{1}} \left(−\mathrm{1}/{a}\right)} ^{{sinh}^{−\mathrm{1}} \left(\mathrm{1}/{a}\right)} −\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{{cos}\left(\mathrm{2}{u}\right)}{\mathrm{2}}+{u}\right]_{{sin}^{−\mathrm{1}} \left(−\mathrm{1}/{a}\right)} ^{{sin}^{−\mathrm{1}} \left(\mathrm{1}/{a}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{a}}\sqrt{\mathrm{1}+\left(\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} }+{sinh}^{−\mathrm{1}} \left(\mathrm{1}/{a}\right)\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{a}}\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{a}^{\mathrm{2}} }}+{sin}^{−\mathrm{1}} \left(\mathrm{1}/{a}\right)\right) \\ $$$${f}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}{{a}^{\mathrm{2}} }+{sinh}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{a}}\right)−\frac{\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}{{a}^{\mathrm{2}} }−{sin}^{−\mathrm{1}} \left(\mathrm{1}/{a}\right)\right) \\ $$$$ \\ $$