Question Number 55615 by Abdo msup. last updated on 28/Feb/19
$${let}\:{F}\left(\alpha\right)=\int_{\alpha} ^{\mathrm{1}+\alpha^{\mathrm{2}} } \:\:\frac{{sin}\left(\alpha{x}\right)}{\mathrm{1}+\alpha{x}^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\frac{{dF}}{{d}\alpha}\left(\alpha\right) \\ $$$$\left.\mathrm{2}\right)\:\:{calculate}\:{lim}_{\alpha\rightarrow\mathrm{0}} \:\:{F}\left(\alpha\right) \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 28/Feb/19
$$\frac{{dF}}{{d}\alpha}=\int_{\alpha} ^{\mathrm{1}+\alpha^{\mathrm{2}} } \frac{\partial}{\partial\alpha}\left(\frac{{sin}\alpha{x}}{\mathrm{1}+\alpha{x}^{\mathrm{2}} }\right){dx}\:+\frac{{sin}\alpha\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}{\mathrm{1}+\alpha\left(\mathrm{1}+\alpha\right)^{\mathrm{2}} }\frac{{d}}{{d}\alpha}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)−\frac{{sin}\alpha\left(\alpha\right)}{\mathrm{1}+\alpha\left(\alpha\right)^{\mathrm{2}} }×\frac{{d}}{{d}\alpha}\left(\alpha\right) \\ $$$$=\int_{\alpha} ^{\mathrm{1}+\alpha^{\mathrm{2}} } \frac{\left(\mathrm{1}+\alpha{x}^{\mathrm{2}} \right)×{cos}\alpha{x}×{x}−{sin}\alpha{x}\left(\mathrm{0}+{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\alpha{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}+\frac{{sin}\left(\alpha+\alpha^{\mathrm{3}} \right)}{\mathrm{1}+\alpha+\mathrm{2}\alpha^{\mathrm{2}} +\alpha^{\mathrm{3}} }×\left(\mathrm{2}\alpha\right)−\frac{{sin}\alpha^{\mathrm{2}} }{\mathrm{1}+\alpha^{\mathrm{3}} }×\mathrm{1} \\ $$$$=\int_{\alpha} ^{\mathrm{1}+\alpha^{\mathrm{2}} } \frac{{xcos}\alpha{x}+\alpha{x}^{\mathrm{3}} {cos}\alpha{x}^{\mathrm{2}} \:−{x}^{\mathrm{2}} {sin}\alpha{x}}{\left(\mathrm{1}+\alpha{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}+\frac{\mathrm{2}\alpha{sin}\left(\alpha+\alpha^{\mathrm{3}} \right)}{\mathrm{1}+\alpha+\mathrm{2}\alpha^{\mathrm{2}} +\alpha^{\mathrm{3}} }−\frac{{sin}\alpha^{\mathrm{2}} }{\mathrm{1}+\alpha^{\mathrm{3}} } \\ $$$${wait}… \\ $$