[m]u=x[/m] ⇒ [m]du=dx[/m]
[m]dv=\frac{dx}{sin^2x}[/m] ⇒ [m]v= ∫\frac{dx}{sin^2x}=-ctgx [/m]
[m]∫_{0}^{\frac{π}{4}} \frac{xdx}{sin^2x}=(x\cdot (-ctgx))|_{0}^{\frac{π}{4}} - ∫ _{0}^{\frac{π}{4}} (-ctgx)dx=[/m]
[m]=(\frac{π}{4}\cdot (-ctg\frac{π}{4}))+lim_{x → 0}\frac{x}{tgx} + ∫ _{0}^{\frac{π}{4}} \frac{cosx}{sinx}dx=[/m]
[m]=(\frac{π}{4}\cdot (-1)+1 +( ln|sinx|)|_{0}^{\frac{π}{4}}=1-\frac{π}{4}+ln|sin \frac{π}{4}|-lim_{x →0} ln|sinx|[/m]
Расходится, так как
[m]lnsinx → ∞ [/m]