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