`int_0^a x^2 sqrt(a^2 - x^2) dx` Evaluate the integral

`int_0^ax^2sqrt(a^2-x^2)dx`

Let's evaluate the indefinite integral by applying integral substitution,

Let `x=asin(u)`

`dx=acos(u)du`

`intx^2sqrt(a^2-x^2)dx=int(asin(u))^2sqrt(a^2-(asin(u))^2)acos(u)du`

`=inta^2sin^2(u)sqrt(a^2-a^2sin^2(u))acos(u)du`

`=a^3intsin^2(u)cos(u)sqrt(a^2(1-sin^2(u)))du`

`=a^3intsin^2(u)cos(u)asqrt(1-sin^2(u))du` 

Now use the identity:`1-sin^2(x)=cos^2(x)`

`=a^4intsin^2(u)cos(u)sqrt(cos^2(u))du`

`=a^4intsin^2(u)cos^2(u)du`

Now use the identity:`cos^2(x)sin^2(x)=(1-cos(4x))/8`

`=a^4int(1-cos(4u))/8du`

`=a^4/8int(1-cos(4u)du`

`=a^4/8(int1du-intcos(4u)du)`

`=a^4/8(u-sin(4u)/4)`

Substitute back `u=arcsin(x/a)`

`=a^4/8(arcsin(x/a)-sin(4arcsin(x/a))/4)`

add a constant C to the solution,

`=a^4/8(arcsin(x/a)-1/4sin(4arcsin(x/a)))+C`

Now let's evaluate the definite integral,

`int_0^ax^2sqrt(a^2-x^2)dx=[a^4/8(arcsin(x/a)-1/4sin(4arcsin(x/a)))]_0^a`

`=[a^4/8(arcsin(a/a)-1/4sin(4arcsin(a/a)))]-[a^4/8(arcsin(0/a)-1/4sin(4arcsin(0/a)))]`

`=[a^4/8(arcsin(1)-1/4sin(4arcsin(1)))]-[a^4/8(arcsin(0)-1/4sin(4arcsin(0)))]`

`=[a^4/8(pi/2-1/4sin(4*pi/2))]-[a4/8(0-1/4sin(4*0))]`

`=[a^4/8(pi/2-1/4sin(2pi))]-[0]`

`=[a^4/8(pi/2-1/4(0))]`

`=a^4/8(pi/2)`

`=(pia^4)/16`