Integration Questions

Question Number 65736 by malwaan last updated on 03/Aug/19

$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\sqrt{\mathrm{1}\:+\:\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} }\:\boldsymbol{{dx}}\:=\:? \\$$

Commented by Souvik Ghosh last updated on 03/Aug/19

$${let}\:\:\:{u}=\mathrm{2}{x}\Leftrightarrow{du}=\mathrm{2}{dx} \\$$$$\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:{du}/\mathrm{2} \\$$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{u}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left\{{u}+\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\right\}\right]_{\mathrm{0}} ^{\mathrm{2}} \\$$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\left[\sqrt{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)\right] \\$$

Commented by mathmax by abdo last updated on 03/Aug/19

$${changement}\:\mathrm{2}{x}\:={sh}\left({t}\right)\:{give}\:{t}\:={argsh}\left(\mathrm{2}{x}\right)\:={ln}\left(\mathrm{2}{x}+\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\right) \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }{dx}\:=\int_{\mathrm{0}} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} \sqrt{\mathrm{1}+{sh}^{\mathrm{2}} {t}}\frac{{ch}\left({t}\right)}{\mathrm{2}}{dt} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} {ch}^{\mathrm{2}} {t}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} \left(\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)\right){dt} \\$$$$=\frac{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{8}}\left[{sh}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} \\$$$$=\frac{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{8}}\left[\frac{{e}^{\mathrm{2}{t}} −{e}^{−\mathrm{2}{t}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} \\$$$$=\frac{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{16}}\left\{\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }\right\} \\$$

Answered by MJS last updated on 03/Aug/19

$$\int\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }{dx}= \\$$$$\:\:\:\:\:\left[{t}=\mathrm{arcsinh}\:\mathrm{2}{x}\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{2}}{dt}\right] \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{cosh}^{\mathrm{2}} \:{t}\:{dt}=\frac{\mathrm{1}}{\mathrm{4}}{t}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sinh}\:\mathrm{2}{t}\:= \\$$$$=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{arcsinh}\:\mathrm{2}{x}\:+\frac{{x}\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{2}}\:+{C} \\$$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{arcsinh}\:\mathrm{2}\:+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\$$$$\left(=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)\:+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\$$