Question Number 192212 by josemate19 last updated on 11/May/23
$$\int−\mathrm{1}^{{x}} {dx} \\ $$
Answered by Frix last updated on 12/May/23
$$\int−\mathrm{1}^{{x}} {dx}=−\int\mathrm{1}^{{x}} {dx}=−\int{dx}=−{x}+{C} \\ $$$$\int\left(−\mathrm{1}\right)^{{x}} {dx}=\int\mathrm{e}^{\mathrm{i}\pi{x}} {dx}=\int\left(\mathrm{cos}\:\pi{x}\:+\mathrm{i}\:\mathrm{sin}\:\pi{x}\right){dx}= \\ $$$$=\frac{\mathrm{sin}\:\pi{x}}{\pi}−\frac{\mathrm{cos}\:\pi{x}}{\pi}\mathrm{i}+{C} \\ $$