int x^2/sqrt(36-x^2) dx Find the indefinite integral

Given
int x^2/sqrt(36-x^2) dx
This can be solved by using the Trigonometric substitutions  (Trig substitutions)
when the integral contains sqrt(a-bx^2) then we have to take
x=sqrt(a/b) sin(t) in order to solve the integral easily
 
so here , For
int x^2/sqrt(36-x^2) dx
x is given as
x= sqrt(36/1) sin(t) = 6sin(t)
=> dx = 6 cos(t) dt
so ,
int x^2/sqrt(36-x^2) dx
=int (6sin(t))^2/sqrt(36-(6sin(t))^2) (6 cos(t) dt)
= int 36(sin(t))^2/sqrt(36-(6sin(t))^2) (6 cos(t) dt)
= int ((36)*(6)(sin(t))^2 *cos(t)) /sqrt(36-(6sin(t))^2) dt
=int (216(sin(t))^2 *cos(t)) /sqrt(36-36(sin(t))^2) dt
= int (216(sin(t))^2 *cos(t)) /sqrt(36(1-(sin(t))^2)) dt
=int (216(sin(t))^2 *cos(t)) /sqrt(36(cos(t))^2) dt
=int (216(sin(t))^2 *cos(t)) /(6(cos(t))) dt
= int (216/6) sin^2(t) dt
= int 36 sin^2(t) dt
= 36 int sin^2(t) dt
= 36 int (1-cos(2t))/2 dt
= (36/2) int (1-cos(2t)) dt
= 18 [int 1 dt - int cos(2t) dt]+c
= 18[t- (1/2)sin(2t)]+c
but we know that
x= 6sin(t)
=> x/6 = sin (t)
=> t= sin^(-1) (x/6) or arcsin(x/6)
so,
18[t- (1/2)sin(2t)]+c
= 18[(arcsin(x/6))- (1/2)sin(2(arcsin(x/6)))]+c
so,
int x^2/sqrt(36-x^2) dx
=18arcsin(x/6)- 9sin(2(arcsin(x/6)))+c