$\int \ x\sqrt{1-x^2}\,dx$, by the substitution $x= \cos t$ I have tried to determine $\int \ x\sqrt{1-x^2}\,dx$ using trigonometric 
formula by the substition $x=\cos t$ I have got :
$$-\int \cos t \sin^2 t\,dt\tag{1}$$ for $\sin t > 0$ and $$ \int \cos t \sin^2 t\,dt\tag{2}$$  for $\sin t <0 $.
But both $(1)$ and $(2)$ have no standard mathematical function , then how do i can determine the titled integral using this method ?

From $\int \cos t \sin^2 t dt$ you can use $u=\sin t, du=\cos t \ dt$ to get $\int u^2 du=\frac {u^3}3+C$ and backsubstitute.
