Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 29

Take the integral int x sqrt(1 - x^4) dx
substitute u = x^2
u = x^2
du = 2x dx
to get 1/2 int sqrt(1 - u^2) du
now make u = sin(t) so:
du = cos(t)dt
1/2 int sqrt(1 - sin^2(t)) cos(t) dt
and since 1 - sin^2(t) is the same thing as cos^2(t) this can be simplified to:
1/2 int cos^2(t) dt
This can be written as
1/2 int 1/2 cos(2t) + 1/2 dt
taking the integral of this we get
1/2 (1/4 sin(2t) + 1/2 t) + c
or 1/8 sin(2t) + 1/4 t
in order to get this back to using x, we need to get sin2t to be in terms of just sin(t)
We can use sin(2t) = 2sin(t)cos(t) and the fact that cos(t) = sqrt(1 - sin^2(t))
to get:
t/4 + 1/4 sin(t) sqrt(1 - sin^2(t))
using u = sin(t), and t = sin^-1(u) we get
(sin^-1(u))/4 + 1/4 u sqrt(1 - u^2)
now sub back in x^2 = u
(sin^-1(x^2))/4 + 1/4 x^2 sqrt(1 - x^4)
This is the final answer.
int x sqrt(1 - x^4) dx =(sin^-1(x^2))/4 + 1/4 x^2 sqrt(1 - x^4)