A cura di: Gianluca
$int1/((sqrt(x))(sqrt(1-x)))dx$
Pongo $sqrt(1-x)=t$ $rarr$$1-x=t^2$ $rarr$ $x=1-t^2$ dx=2tdt
$int1/(sqrt(1-t^2))[1/t(2t)dt]$
=$2int1/(sqrt(1-t^2))dt$
=$2arcsensqrt(x)+C$
di Anoè Gianluca – www.chenesoio.com
- Integrali
- Matematica - Integrali
