[m] ∫_{\frac{π}{4}}^{\frac{π}{2}}\frac{dx}{sin^2x}=(-ctgx)|_{\frac{π}{4}}^{\frac{π}{2}}=-ctg \frac{π}{2}+ctg\frac{π}{4}=-0+1=1 [/m]
[m] ∫_{1}^{2}(1-\frac{1}{x^2})dx=(x-\frac{x^{-2+1}}{-2+1}) _{1}^{2}=(x+\frac{1}{x}) _{1}^{2}=(2-1)+(\frac{1}{2}-1)=\frac{1}{2}[/m]
Обсуждения