Question

решите пожалуйста. Очень надо.

Answer 1

$\int _1^2 \frac{1}{x^4}dx=\int _1^2 x^{-4}dx=\frac{x^{-3}}{-3}|_1^2=-\frac{1}{3x^3}|_1^2=-\frac{1}{24}+\frac{1}{3}=\frac{7}{24}\\\\\int _\frac{\pi}{2}^\frac{\pi}{4}\frac{2}{sin^2x}dx=-2\cdot ctgx|_{\frac{\pi}{2}}^{\frac{\pi}{4}}=-2(ctg\frac{\pi}{4}-ctg\frac{\pi}{2})=-2(1-0)=-2$