Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 21

int_0^0.6x^2/sqrt(9-25x^2)dx
Let's first evaluate the indefinite integral by applying the integral substitution,
Let x=3/5sin(u)
=>dx=3/5cos(u)du
Plug in the above in the integral,
intx^2/sqrt(9-25x^2)dx=int(3/5sin(u))^2/(sqrt(9-25(3/5sin(u))^2))(3/5cos(u))du
=int(27sin^2(u)cos(u))/(125sqrt(9-9sin^2(u)))du
=int(27sin^2(u)cos(u))/(125sqrt(9(1-sin^2(u))))du
use the identity:1-sin^2(x)=cos^2(x)

=int(27sin^2(u)cos(u))/(125*3sqrt(cos^2(u)))du
=int(9sin^2(u)cos(u))/(125cos(u))du
=9/125intsin^2(u)du
Now use the identity:sin^2(x)=(1-cos(2x))/2
=9/125int(1-cos(2u))/2du
=9/250int(1-cos(2u))du
=9/250(int1du-intcos(2u)du)
=9/250(u-sin(2u)/2)
We have taken x=3/5sin(u)
=>u=arcsin(5/3x)
Substitute back u and add a constant C to the solution,
=9/250(arcsin(5/3x)-1/2sin(2arcsin(5/3x)))+C
Now let's evaluate the definite integral,
int_0^0.6x^2/sqrt(9-25x^2)dx=[9/250(arcsin(5/3x)-1/2sin(2arcsin(5/3x)))]_0^0.6
=[9/250(arcsin(5/3*0.6)-1/2sin(2arcsin(5/3*0.6)))]-[9/250(arcsin(5/3*0)-1/2sin(2arcsin(5/3*0)))]
=[9/250(arcsin(1)-1/2sin(2arcsin(1)))]-[9/250(arcsin(0)-1/2sin(2arcsin(0)))]
=[9/250(pi/2-1/2sin(2*pi/2))]-[0]
=[9/250(pi/2-1/2sin(pi))]
=[9/250(pi/2-1/2*0)]
=(9pi)/500
=0.05655