Question Number 60311 by aliesam last updated on 19/May/19
$$\int\frac{{dx}}{\:\sqrt{{sec}\:{h}^{\mathrm{2}} \left({x}\right)+\mathrm{1}}}\:{dx} \\ $$
Commented by MJS last updated on 20/May/19
$$\mathrm{sech}\:{x}=\frac{\mathrm{1}}{\mathrm{cosh}\:{x}} \\ $$
Commented by maxmathsup by imad last updated on 20/May/19
$${what}\:{is}\:{sech}\left({x}\right)?…{i}\:{dont}\:{use}\:{this}\:{notation}… \\ $$
Answered by MJS last updated on 20/May/19
![∫(dx/( (√(1+sech^2 x))))=∫((cosh x)/( (√(1+cosh^2 x))))dx= [t=sinh x → dx=(dt/(cosh x))] =∫(dt/( (√(t^2 +2))))=ln (t+(√(t^2 +2)))= =ln (sinh x +(√(2+sinh^2 x))) +C](https://www.tinkutara.com/question/Q60360.png)
$$\int\frac{{dx}}{\:\sqrt{\mathrm{1}+\mathrm{sech}^{\mathrm{2}} \:{x}}}=\int\frac{\mathrm{cosh}\:{x}}{\:\sqrt{\mathrm{1}+\mathrm{cosh}^{\mathrm{2}} \:{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{sinh}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{cosh}\:{x}}\right] \\ $$$$=\int\frac{{dt}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{2}}}=\mathrm{ln}\:\left({t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{2}}\right)= \\ $$$$=\mathrm{ln}\:\left(\mathrm{sinh}\:{x}\:+\sqrt{\mathrm{2}+\mathrm{sinh}^{\mathrm{2}} \:{x}}\right)\:+{C} \\ $$
Commented by aliesam last updated on 20/May/19
$$\int\frac{{dx}}{\:\sqrt{\mathrm{1}+{sech}^{\mathrm{2}} \left({x}\right)}}=\int\frac{{cosh}\left({x}\right)}{\:\sqrt{\mathrm{1}+{cosh}^{\mathrm{2}} \left({x}\right)}} \\ $$
Commented by MJS last updated on 20/May/19
$$\mathrm{yes}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{a}\:\mathrm{typo} \\ $$
Commented by aliesam last updated on 20/May/19
$${yes}.{thank}\:{you}\:{zir} \\ $$