Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 16

You need to use integration by parts, such that:
int u dv = uv - int vdu
u = x => du = dx
dv = sin^3 x => v = int sin^3 x dx
You need to solve the integral int sin^3 x dx such that:
int sin^3 x dx = int sin^2 x* sin xdx
Replace 1 - cos^2 x for sin^2 x , such that:
int sin^2 x* sin xdx = int (1 - cos^2 x)* sin xdx
Use substitution cos x = t => -sin x dx = dt:
int (1 - cos^2 x)* sin xdx = int (1 - t^2)* (-dt)
int (1 - t^2)* (-dt) = int (t^2 - 1) dt
int (t^2 - 1) dt = int t^2 dt - int dt
int (t^2 - 1) dt = t^3/3 - t + c
Replace back cos x for t:
int (1 - cos^2 x)* sin x dx = (cos^3 x)/3 - cos x + c
Hence, v = int sin^3 x dx = (cos^3 x)/3 - cos x
Using parts, yields:
int x sin^3 x dx = x*((cos^3 x)/3 - cos x) - int ((cos^3 x)/3 - cos x) dx
int x sin^3 x dx = x*((cos^3 x)/3 - cos x) - int (cos^3 x)/3 dx + int cos x dx
You need to solve the integral int (cos^3 x)/3dx , such that:
int (cos^3 x)/3dx = (1/3) int cos^2 x*cos x dx
Replace 1 - sin^2 x for cos^2 x:
(1/3) int (1 - sin^2 x)*cos x dx
Use substitution sin x = t => cos x dx = dt
(1/3) int (1 - sin^2 x)*cos x dx = (1/3)*int (1 - t^2) dt = (1/3) t - (1/3)(t^3)/3 + c
(1/3) int (1 - sin^2 x)*cos x dx = (1/3) sin x- (1/3)(sin^3 x)/3 + c
Hence, the result of integration is: int x sin^3 x dx = x*((cos^3 x)/3 -cos x) - ((1/3) sin x- (1/3)(sin^3 x)/3) + sin x + c