`int x sin^3(x) dx` Evaluate the integral

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`