Let f(x) = sin^3 x

We need to find the integral of f(x).

==> intg f(x) = intg sin^3 x dx

= intg sin^2 x * sin x dx

= intg ( 1- cos^2 x) * sinx dx

Let us assume that u = cosx ==> du = -sin x dx

Let us substitute:

==> intg f(x) = intg ( 1- u^2) * sinx * du/-sinx

= -intg - (1- u^2) du

= intg ( u^2 - 1) du

= u^3/3 - u + C

Now we will substitutee with u = cosx

**==> intg f(x) = cos^3 x/ 3 - cosx + C**

**==> intg sin^3 x = (1/3)*cos^3 x - cosx + C**

Int (sin x)^3dx

We'll write (sin x)^3 = (sin x)^2*(sin x)

We'll re-write the integral:

Int (sin x)^3dx = Int (sin x)^2*(sin x)dx

From the fundamental formula of trigonometry, we'll get:

(sin x)^2 = 1 - (cos x)^2

We'll re-write the integral:

Int [1 - (cos x)^2]*(sin x)dx

We'll substitute cos x = t.

We'll differentiate both sides:

-sin xdx = dt

We'll re-write the integral in t:

- Int (1 - t^2)dt = -Int dt + Int t^2

**Int (sin x)^3dx = (cos x)^3/3 - cos x + C**