Задача 59042 ...
Условие
математика колледж
196
Решение
★
u=arcsin(1-x)
du=(arcsin(1-x))`dx=[m]\frac{1}{\sqrt{1-(1-x)^2}}\cdot (1-x)`dx=-\frac{1}{\sqrt{1-(1-x)^2}}dx[/m]
dv=dx
v=x
[m] ∫ ^{1}_{\frac{1}{2}} arcsin(1-x)dx=x\cdot arcsin(1-x)|^{1}_{\frac{1}{2}} - ∫ ^{1}_{\frac{1}{2}} (-\frac{1}{\sqrt{1-(1-x)^2}})dx=[/m]
[m]=1\cdot arcsin(1-1)-\frac{1}{2}\cdot arcsin(1-\frac{1}{2})-∫ ^{1}_{\frac{1}{2}}\frac{1}{\sqrt{1-(1-x)^2}})d(1-x)=[/m]
[m]=1\cdot 0-\frac{1}{2}\cdot arcsin\frac{1}{2}-arcsin(1-x)|^{1}_{\frac{1}{2}}=[/m]
[m]=-\frac{1}{2}\cdot\frac{π}{6}-arcsin(1-1)+arcsin(1-\frac{1}{2})=-\frac{π}{12}+\frac{π}{6}=\frac{π}{12}[/m]