int 4arccosx dx Find the indefinite integral

Given ,
y= int 4 cos^(-1)x dx
By Applying the integration by parts we get,
this solution
so,
let u=cos^(-1)x=> u'= (cos^(-1)x )'
as we know (d/dx)cos^(-1)x =(-1)/(sqrt(1-x^2))
and v'=1 =>v =x
now by Integration by parts ,
int uv' dx= uv-int u'v dx
so , now
int (cos^(-1)x )dx
= (cos^(-1)x )(x) - int ((-1)/(sqrt(1-x^2)) )*x dx
= x(cos^(-1)x ) + int ((x)/(sqrt(1-x^2)) ) dx
let 1-x^2 = q
=> -2x dx= dq
so ,
int ((x)/(sqrt(1-x^2)) )dx
=(-1/2)int ((-2x)/(sqrt(1-x^2)) )dx
= (-1/2)int (1/(sqrt(q))) dq = (-1/2)q^((-1/2)+1)/((-1/2)+1) =-sqrt(q) = - sqrt(1-x^2)
 so ,now
int (cos^(-1)x )dx
= (cos^(-1)x )(x)  -(sqrt(1-x^2))
and now
int 4(cos^(-1)x )dx
=4int (cos^(-1)x )dx
=4(x(cos^(-1)x )  -(sqrt(1-x^2)))+c