Question Number 83205 by Rio Michael last updated on 28/Feb/20
$$\int_{\mathrm{0}} ^{{ln}\mathrm{2}} \frac{\mathrm{1}}{{cosh}\left({x}\:+\:{ln}\mathrm{4}\right)}{dx}\:= \\ $$
Commented by mathmax by abdo last updated on 28/Feb/20
![I =_(x+ln4=t) ∫_(2ln2) ^(ln(6)) (dt/(ch(t))) =∫_(2ln2) ^(ln6) ((2dt)/(e^t +e^(−t) )) =_(e^t =u) ∫_4 ^6 ((2du)/(u(u+u^(−1) ))) =2∫_4 ^6 (du/(u^2 +1)) =2[arctan(u)]_4 ^6 =2(arctan(6)−arctan(4))](https://www.tinkutara.com/question/Q83217.png)
$${I}\:=_{{x}+{ln}\mathrm{4}={t}} \:\:\:\int_{\mathrm{2}{ln}\mathrm{2}} ^{{ln}\left(\mathrm{6}\right)} \:\frac{{dt}}{{ch}\left({t}\right)}\:=\int_{\mathrm{2}{ln}\mathrm{2}} ^{{ln}\mathrm{6}} \:\frac{\mathrm{2}{dt}}{{e}^{{t}} \:+{e}^{−{t}} } \\ $$$$=_{{e}^{{t}} ={u}} \:\:\:\:\:\int_{\mathrm{4}} ^{\mathrm{6}} \:\:\frac{\mathrm{2}{du}}{{u}\left({u}+{u}^{−\mathrm{1}} \right)}\:=\mathrm{2}\int_{\mathrm{4}} ^{\mathrm{6}} \:\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:=\mathrm{2}\left[{arctan}\left({u}\right)\right]_{\mathrm{4}} ^{\mathrm{6}} \\ $$$$=\mathrm{2}\left({arctan}\left(\mathrm{6}\right)−{arctan}\left(\mathrm{4}\right)\right) \\ $$
Commented by Rio Michael last updated on 28/Feb/20
$${thanks}\:{sir} \\ $$