A cura di: Francesco Speciale

Svolgimento:
Ponendo$t=cosx$, risulta $dx=-(dt)/(sqrt(1-t^2))$.Perciò:
$int_(0)^((pi)/2)(sqrt(1+cosx))dx=-int_(0)^(1)((sqrt(1+t))/(sqrt(1-t^2)))dt=$
$int_(0)^(1)(1/(sqrt(1-t)))dt=[2sqrt(1-t)]_(0)^(1)=2$.