Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 15

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