We need the integral of f(x) = (sin 5x)^3

(sin 5x)^3 = sin 5x * (sin 5x)^2

=> sin 5x ( 1 - (cos 5x)^2)

Now let cos 5x = t

dt/dx = -5 sin 5x

Int [ sin 5x ( 1 - (cos 5x)^2) dx]

=> Int [ (-1/5) ( 1 - t^2) dt]

=> (-1/5)[ t - (t^3)/3] + C

=> -t/5 + t^3 / 5*3 + C

substitute t = cos 5x

=> -(cos 5x)/5 + ( cos 5x)^3/15 +C

**Therefore the integral of f(x) = (sin 5x)^3 is -(cos 5x)/5 + ( cos 5x)^3/15 + C**

To find the antiderivative of f(x), we'll ahve to evaluate the indefinite integral of f(x).

Int f(x)dx = F(x) + C, where F(x) is the antiderivative of f(x).

We'll apply recursion formula to calculate:

Int [sin (5x)]^3dx

We know that:

Int (sin u)^n du = (-1/n)/(sin u)^(n-1)*(cos u) + [(n-1)/n]*Int (sin u)^(n-2) du

We'll substitute 5x = t.

We'll differentiate both sides:

5dx = dt

dx = dt/5

Int [sin (5x)]^3dx = (1/5) Int (sin t)^3 dt

According to the rule:

Int (sin t)^3 dt = (-1/3)(sin t)^2*(cos t) + (2/3) Int sin t dt

Int (sin t)^3 dt = (-1/3)(sin t)^2*(cos t) - (2/3)*(cos t)

**Int [sin (5x)]^3dx = -(sin 5x)^2*(cos 5x) /15 - 2cos 5x/15 + C**