How to find out the indefinite integral of the cube of sinx?

`int` (sinx)^3dx = `int` [(sinx)^2*sin
x]dx

According to Pythagorean identity, we'll
get:

(sinx)^2  = 1 -
(cosx)^2

`int` [(sinx)^2*sin x]dx = `int` [(1 -
(cosx)^2)*sin x]dx

We'll remove the
brackets:

`int` [(1 - (cosx)^2)*sin x]dx  = `int` sin xdx -
`int` (cosx)^2*sin xdx

We'll evaluate `int` (cosx)^2*sin
xdx using substitution:

cos x =
t

We'll differentiate both
sides:

-sin x dx = dt

We'll
re-write the integral, changing the variable:

`int`
(cosx)^2*sin xdx = -`int` t^2dt

-`int` t^2dt = -t^3/3 +
C

`int` (cosx)^2*sin xdx = -(cos x)^3/3 +
C

`int` (sinx)^3dx = `int` sin xdx - `int` (cosx)^2*sin
xdx

`int` (sinx)^3dx = -cos x + (cos x)^3/3 +
C