Question Number 149329 by mnjuly1970 last updated on 04/Aug/21
Commented by Kamel last updated on 04/Aug/21
![I think I^(100) +k≡0[14]](https://www.tinkutara.com/question/Q149366.png)
$${I}\:{think}\:{I}^{\mathrm{100}} +{k}\equiv\mathrm{0}\left[\mathrm{14}\right] \\ $$
Answered by Kamel last updated on 04/Aug/21
![∫_0 ^(π/2) ((e^(−tan(x)) sin^2 (x))/(cos^6 (x)))dx=^(t=tan(x)) ∫_0 ^(+∞) (t^2 +t^4 )e^(−t) dt=2+4!=26 26^(100) +k≡0[14] 26≡−2[14]⇒26^(100) ≡2^(100) [14] 2^4 ≡2[14]⇒2^(100) ≡2^(25) [14]⇒2^(100) ≡2^(10) [14]⇒2^(100) ≡2[14] ∴ k≡−26^(100) [14]⇒k≡−2[14]⇒k≡12[14] ∴ k=14n+12⇒k_(min) =12, k>0.](https://www.tinkutara.com/question/Q149365.png)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{e}^{−{tan}\left({x}\right)} {sin}^{\mathrm{2}} \left({x}\right)}{{cos}^{\mathrm{6}} \left({x}\right)}{dx}\overset{{t}={tan}\left({x}\right)} {=}\int_{\mathrm{0}} ^{+\infty} \left({t}^{\mathrm{2}} +{t}^{\mathrm{4}} \right){e}^{−{t}} {dt}=\mathrm{2}+\mathrm{4}!=\mathrm{26} \\ $$$$\mathrm{26}^{\mathrm{100}} +{k}\equiv\mathrm{0}\left[\mathrm{14}\right] \\ $$$$\mathrm{26}\equiv−\mathrm{2}\left[\mathrm{14}\right]\Rightarrow\mathrm{26}^{\mathrm{100}} \equiv\mathrm{2}^{\mathrm{100}} \left[\mathrm{14}\right] \\ $$$$\mathrm{2}^{\mathrm{4}} \equiv\mathrm{2}\left[\mathrm{14}\right]\Rightarrow\mathrm{2}^{\mathrm{100}} \equiv\mathrm{2}^{\mathrm{25}} \left[\mathrm{14}\right]\Rightarrow\mathrm{2}^{\mathrm{100}} \equiv\mathrm{2}^{\mathrm{10}} \left[\mathrm{14}\right]\Rightarrow\mathrm{2}^{\mathrm{100}} \equiv\mathrm{2}\left[\mathrm{14}\right] \\ $$$$\therefore\:{k}\equiv−\mathrm{26}^{\mathrm{100}} \left[\mathrm{14}\right]\Rightarrow{k}\equiv−\mathrm{2}\left[\mathrm{14}\right]\Rightarrow{k}\equiv\mathrm{12}\left[\mathrm{14}\right] \\ $$$$\therefore\:{k}=\mathrm{14}{n}+\mathrm{12}\Rightarrow{k}_{{min}} =\mathrm{12},\:{k}>\mathrm{0}. \\ $$
Commented by mnjuly1970 last updated on 05/Aug/21
$$\:\:\:\:\:\:\:\:\:{tashakor}\:\:{mr}\:{kamel}…. \\ $$