Question

интеграл от 0 до п/2  от sin^3xdx

Answer 1

$intlimits_0^{fracpi2}sin^3xdx=intlimits_0^{fracpi2}sin^2xcdotsin xdx=-intlimits_0^{fracpi2}sin^2x dcos x=intlimits_{fracpi2}^0sin^2x dcos x=\ =intlimits_{fracpi2}^0(1-cos^2x) dcos x=intlimits_{fracpi2}^0 dcos x-intlimits_{fracpi2}^0cos^2x dcos x=\ =cos x|_fracpi2^0-frac{1}{3}cos^3x|_fracpi2^0=(cos0-cosfracpi2)-frac13(cos^30-cos^3fracpi2)=\ =(1-0)-frac13(1-0)=1-frac13=frac23$

Answer 2

спасибо