Question

integrală prin schimbare de variabilă.. vreo idee?

Answer 1

arcsinx=u     dx/√(1-x²)=du
∫√udu=∫u^(1/2)du=u^(3/2)/(3/2)=2u^(3/2)/3
Se  revine  la  variabila  x
∫...=3/2*(arcsinx)^(3/2)+c
S-a  folosit  formula
∫u^ndu=u^(n+1)/du+C

Answer 2

$I= \int\limits { \sqrt{ \frac{arcsinx}{1- x^{2} } } } \, dx$$= \int\limits { \frac{ \sqrt{arcsinx} }{ \sqrt{1- x^{2} } } } \, dx$
Notam $\sqrt{arcsinx}=t,deci,arcsinx=t^2,derivand, \frac{1}{ \sqrt{1- x^{2} } }dx=2tdt,$. Integrala devine: 
$I=2 \int\limits {t^2} \, dt= \frac{2}{3}t^3+C= \frac{2}{3} \sqrt{arcsinx}^3+C$