Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 4

You need to use the following substitution, such that:
x = sin t => dx = cos t dt
1 - x^2 = 1 - sin^2 t = cos^2 t
Changing the variable, yields:
int_0^1 x^3*sqrt(1 - x^2)dx = int_(t_1)^(t_2) sin^3 t*sqrt (cos^2 t)*cos t dt
int_(t_1)^(t_2) sin^3 t*sqrt (cos^2 t)*cos t dt = int_(t_1)^(t_2) sin^3 t*cos t*cos t dt
int_(t_1)^(t_2) sin^3 t*cos^2 t dt = int_(t_1)^(t_2) sin^2 t*cos^2 t*sin t dt
int_(t_1)^(t_2) sin^2 t*cos^2 t*sin tdt = int_(t_1)^(t_2) (1 - cos^2 t)*cos^2 t*sin tdt
You need to make the next substitution, such that:
cos t = u => -sin t dt = du
int_(u_1)^(u_2) (1 - u^2)*u^2*(-du) =- int_(u_1)^(u_2)u^2 du + int_(u_1)^(u_2)u^4du
int_(u_1)^(u_2) (1 - u^2)*u^2*(-du) =(-(cos t)^3/3 + (cos t)^5/5)|_(t_1)^(t_2)
Since x = sin t => t = arcsin x
int_0^1 x^3*sqrt(1 - x^2)dx = (-(cos (arcsin x))^3/3 + (cos (arcsin x)^5/5)|_(0)^(1)
int_0^1 x^3*sqrt(1 - x^2)dx = (-(cos (arcsin 1))^3/3 + (cos (arcsin 1)^5/5 + (cos (arcsin 0))^3/3 - (cos (arcsin 0))^5/5)
int_0^1 x^3*sqrt(1 - x^2)dx = -(cos (pi/2))^3/3 + (cos(pi/2))^5/5 + (cos 0)^3/3 - (cos 0)^5/5
int_0^1 x^3*sqrt(1 - x^2)dx = 1/3 - 1/5
int_0^1 x^3*sqrt(1 - x^2)dx = 2/15
Hence, evaluating the definite integral yields int_0^1 x^3*sqrt(1 - x^2)dx = 2/15.