How to integrate the following integral using trigonometric substitution? indefinite intergal x^2 /sqr ( 4 -x^2) dx

Int x^2/ sqrt(4-x^2) dx

To
use trigonometric substitution we follow the following
identities:

We know
that:

sqrt(a^2- x^2) ==> x =
a*sin(t).............(1)

sqr(a^2+x^2) ==> x =
a*tan(t)...............(2)

sqrt(x^2-a^2) ==> x =
a*sec(t)................(3)

Comparing the given problem, we
can apply the identity number (1).

==> sqrt(4-x^2)
==> x = 2*sin(t)  ==> dx = 2cos(t)
dt

==> x^2 = 4sin^2
(t)

Now we will
substitute:

==> Int (4sin^2 (t)/sqrt(4-4sin^2 t)
*2cos(t) dt

==> Int (8sin^2 t*cos(t) /sqrt(4(1-sin^2
t)   dt

==> 8 Int (sin^2 t cos(t) / 2sqrt(cos^2 t) 
dt

==> 8 Int sin^2 t cos (t) / 2cos(t)
dt

==> 4 Int sin^2 t
dt

Now we know that sin^2 t =
(1-cos2t)/2

==> 4 Int (1-cos2t)/2 
dt

==> 2 Int (1- cos2t)
dt

==> 2 [ t - sin2t/2 ] +
C

==> 2t - sin2t +
C

Now we will substitute with x = 2sint ==> sint =
x/2 ==> t= arcsin(x/2)

==> 2(arcsin(x/2) -
sin2t + C...........(i)

Since x=
2sint

sin2t = 2sint *cost ==> sin2t =
x*cost

cost = sqr(1-sin^2t) = sqrt(1- (x/2)^2) = 
sqrt(1-x^2/4)= sqrt(4-x^2)/4 = (1/2)sqrt(4-x^2)

==>
sin2t = 2sint*cost = x*(1/2)*sqrt(4-x^2)

==> sin2t =
(1/2)*x*sqrt(4-x^2)

Substitute into (i)
:

==> Int x^2/sqrt(4-x^2) =
2arcsin(x/2) -(1/2)x*sqrt(4-x^2) + C