Question Number 149329 by mnjuly1970 last updated on 04/Aug/21
Commented by Kamel last updated on 04/Aug/21
$${I}\:{think}\:{I}^{\mathrm{100}} +{k}\equiv\mathrm{0}\left[\mathrm{14}\right] \\ $$
Answered by Kamel last updated on 04/Aug/21
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{e}^{−{tan}\left({x}\right)} {sin}^{\mathrm{2}} \left({x}\right)}{{cos}^{\mathrm{6}} \left({x}\right)}{dx}\overset{{t}={tan}\left({x}\right)} {=}\int_{\mathrm{0}} ^{+\infty} \left({t}^{\mathrm{2}} +{t}^{\mathrm{4}} \right){e}^{−{t}} {dt}=\mathrm{2}+\mathrm{4}!=\mathrm{26} \\ $$$$\mathrm{26}^{\mathrm{100}} +{k}\equiv\mathrm{0}\left[\mathrm{14}\right] \\ $$$$\mathrm{26}\equiv−\mathrm{2}\left[\mathrm{14}\right]\Rightarrow\mathrm{26}^{\mathrm{100}} \equiv\mathrm{2}^{\mathrm{100}} \left[\mathrm{14}\right] \\ $$$$\mathrm{2}^{\mathrm{4}} \equiv\mathrm{2}\left[\mathrm{14}\right]\Rightarrow\mathrm{2}^{\mathrm{100}} \equiv\mathrm{2}^{\mathrm{25}} \left[\mathrm{14}\right]\Rightarrow\mathrm{2}^{\mathrm{100}} \equiv\mathrm{2}^{\mathrm{10}} \left[\mathrm{14}\right]\Rightarrow\mathrm{2}^{\mathrm{100}} \equiv\mathrm{2}\left[\mathrm{14}\right] \\ $$$$\therefore\:{k}\equiv−\mathrm{26}^{\mathrm{100}} \left[\mathrm{14}\right]\Rightarrow{k}\equiv−\mathrm{2}\left[\mathrm{14}\right]\Rightarrow{k}\equiv\mathrm{12}\left[\mathrm{14}\right] \\ $$$$\therefore\:{k}=\mathrm{14}{n}+\mathrm{12}\Rightarrow{k}_{{min}} =\mathrm{12},\:{k}>\mathrm{0}. \\ $$
Commented by mnjuly1970 last updated on 05/Aug/21
$$\:\:\:\:\:\:\:\:\:{tashakor}\:\:{mr}\:{kamel}…. \\ $$