Question

2.5 посмотрите пожалуйста

Answer 1

$intlimits_{-frac12}^{frac12}arccos2xdx=left(begin{array}{c}u=2x\du=2dxend{array}right)=frac12intlimits_{-frac12}^{frac12}arccos udu=\=frac12left.left(uarccos u-sqrt{1-u^2}right)right|limits_{-frac12}^{frac12}=left(begin{array}{c}u=2xend{array}right)=\=left.frac12(2xarccos2x-sqrt{1-(2x)^2})right|limits_{-frac12}^{frac12}=left.(xarccos2x-frac12sqrt{1-4x^2})right|limits_{-frac12}^{frac12}=$
$=(frac12arccos1-frac12sqrt0)-(-frac12arccos(-1)-frac12sqrt0)=\=(0-0)-(-frac12cdotpi-0)=fracpi2$