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Question Number 216670 by ahmed2025 last updated on 15/Feb/25

Answered by shunmisaki007 last updated on 15/Feb/25

$$\int {\cos }^{-x}\left(\pi \right)dx=\int {(-1)}^{-x}dx$$$$=\int {\left(e^{\pi i}\right)}^{-x}dx=\int e^{-\pi ix}dx$$$$=\frac{1}{-\pi i}{e}^{-\pi ix}+C$$$$\therefore \int {\cos }^{-x}\left(\pi \right)dx=\frac{i{(-1)}^{-x}}{\pi }+C★$$

Answered by MrGaster last updated on 15/Feb/25

$$\text{Missing left or extra right}$$

Answered by Ghisom last updated on 16/Feb/25

$$\int {\left(\cos \pi \right)}^{-x}dx=\int {(-1)}^{-x}dx=$$$$=\int (\cos \pi x-\mathrm{i}\sin \pi x)dx=$$$$=\frac{1}{\pi }(\sin \pi x+\mathrm{i}\cos \pi x)+C$$

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