Integration Questions

Question Number 35379 by lilesh7 last updated on 18/May/18

$$\int_{\mathrm{0}} ^{\:\:\mathrm{1}} {t}^{\mathrm{2}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:=\:? \\$$

Commented by prof Abdo imad last updated on 18/May/18

$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}} \:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}.{changement}\:{t}={shx} \\$$$${give}\:{I}\:=\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\left.\mathrm{2}\right)}\right.} \:\:{sh}^{\mathrm{2}} {x}\:{chx}.{chxdx} \\$$$$=\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\:\left({shx}\:{chx}\right)^{\mathrm{2}} {dx} \\$$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} {sh}^{\mathrm{2}} \left(\mathrm{2}{x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\frac{{ch}\left(\mathrm{4}{x}\right)−\mathrm{1}}{\mathrm{2}}{dx} \\$$$$=\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} {ch}\left(\mathrm{4}{x}\right){dx}\:−\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\$$$$=\:\frac{\mathrm{1}}{\mathrm{32}}\left[\:{sh}\left(\mathrm{4}{x}\right)\right]_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\:−\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\$$$$=\frac{\mathrm{1}}{\mathrm{32}}\:{sh}\left(\mathrm{4}{ln}\left(\mathrm{1}+\sqrt{\left.\mathrm{2}\right)}\right)\:−\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:{but}\:{we}\:{have}\right. \\$$$${shx}\:=\frac{{e}^{{x}} \:\:−{e}^{−{x}} }{\mathrm{2}}\:\Rightarrow{sh}\left(\mathrm{4}{ln}\left(\mathrm{1}+\sqrt{\left.\mathrm{2}\left.\right)\right)}\right.\right. \\$$$$=\frac{{e}^{\mathrm{4}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:−\:{e}^{−\mathrm{4}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} }{\mathrm{2}}=\frac{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{4}} \:\:−\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{4}} }}{\mathrm{2}}\Rightarrow \\$$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{64}}\left\{\:\:\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{4}} \:−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{−\mathrm{4}} \right\}\:−\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\$$$${let}\:{remember}\:{that} \\$$$${argsh}\left({x}\right)={ln}\left({x}+\sqrt{\left.{x}^{\mathrm{2}} +\mathrm{1}\right)}\right. \\$$$${argchx}={ln}\left({x}\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right) \\$$$$\\$$

Commented by ajfour last updated on 18/May/18

$${I}=\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{8}}−\frac{\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{8}}\:. \\$$$${is}\:{it}\:{same}\:{as}\:{your}\:{answer}\:{to}\:{this} \\$$$${question},{Sir}\:? \\$$

Commented by abdo mathsup 649 cc last updated on 18/May/18

$${perhaps}\:{after}\:{develpping}\:{the}\:{calculus}... \\$$

Answered by MJS last updated on 18/May/18

$$\int{t}^{\mathrm{2}} \sqrt{{t}^{\mathrm{2}} +\mathrm{1}}{dt}=\int\left({t}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}{dt}= \\$$$$=\int\left(\left({t}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}−\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\right){dt}= \\$$$$=\int\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} {dt}−\int\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dt}= \\$$$$\:\:\:\:\:\:\:\:\:\:\left[{t}=\mathrm{tan}\left({u}\right)\:\rightarrow\:{dt}=\mathrm{sec}^{\mathrm{2}} \left({u}\right){du}\right] \\$$$$=\int\mathrm{sec}^{\mathrm{2}} \left({u}\right)\left(\mathrm{tan}^{\mathrm{2}} \left({u}\right)+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} {du}−\int\mathrm{sec}^{\mathrm{2}} \left({u}\right)\left(\mathrm{tan}^{\mathrm{2}} \left({u}\right)+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {du}= \\$$$$\:\:\:\:\:\:\:\:\:\:\left[\mathrm{tan}^{\mathrm{2}} \left({u}\right)+\mathrm{1}=\mathrm{sec}^{\mathrm{2}} \left({u}\right)\right] \\$$$$=\int\mathrm{sec}^{\mathrm{5}} \left({u}\right){du}−\int\mathrm{sec}^{\mathrm{3}} \left({u}\right){du} \\$$$$\\$$$$\mathrm{now}\:\mathrm{use}\:\mathrm{the}\:\mathrm{reduction}\:\mathrm{formula} \\$$$$\int\mathrm{sec}^{{n}} \left(\alpha\right){d}\alpha=\frac{\mathrm{sec}^{{n}−\mathrm{2}} \left(\alpha\right)\mathrm{tan}\left(\alpha\right)}{{n}−\mathrm{1}}+\frac{{n}−\mathrm{2}}{{n}−\mathrm{1}}\int\mathrm{sec}^{{n}−\mathrm{2}} \left(\alpha\right){d}\alpha \\$$

Answered by sma3l2996 last updated on 18/May/18

$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} {t}×{t}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\$$$${by}\:{parts} \\$$$${u}={t}\Rightarrow{u}'=\mathrm{1} \\$$$${v}'={t}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\Rightarrow{v}=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{t}^{\mathrm{2}} } \\$$$${So}\:\:\:{I}=\frac{\mathrm{1}}{\mathrm{3}}\left[{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\$$$${I}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }+{t}^{\mathrm{2}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right){dt}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }{dt}−\frac{\mathrm{1}}{\mathrm{3}}{I} \\$$$$\frac{\mathrm{4}}{\mathrm{3}}{I}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\$$$${let}\:\:{t}={sinh}\left({u}\right)\Rightarrow{dt}={cosh}\left({u}\right){du} \\$$$$\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }=\sqrt{\mathrm{1}+{sinh}^{\mathrm{2}} {u}}={cosh}\left({u}\right) \\$$$${So}\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\int_{\mathrm{0}} ^{{sinh}^{−\mathrm{1}} \left(\mathrm{1}\right)} {cosh}^{\mathrm{2}} \left({u}\right){du}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{{sinh}^{−\mathrm{1}} \left(\mathrm{1}\right)} \left(\mathrm{1}+{cosh}\left(\mathrm{2}{u}\right)\right){du} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{u}+\frac{\mathrm{1}}{\mathrm{2}}{sinh}\left(\mathrm{2}{u}\right)\right]_{\mathrm{0}} ^{{sinh}^{−\mathrm{1}} \left(\mathrm{1}\right)} =\frac{\mathrm{1}}{\mathrm{2}}\left({sinh}^{−\mathrm{1}} \left(\mathrm{1}\right)+\sqrt{\mathrm{2}}\right) \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)+\sqrt{\mathrm{2}}\right) \\$$$${So}\:\:{I}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right) \\$$$${I}=\frac{\mathrm{7}\sqrt{\mathrm{2}}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\$$