y= cos^3 x

Let us rewrite:

y= cos^2 x * cosx

y= ( 1- sin^2 x) cosx

==> y= cosx - sin^2 x * cosx

==> Int y= Int cosx - Int sin^2 x*cosx dx

==> Int y= sinx - Int sin^2 x cosx dx

Let u= sinx ==> u' = cosx dx

==> Int sin^2 x cosx dx = Int u^2 * cosx * du/cosx

==> Int sin^2 x cosx dx = Int u^2 du = u^3/3

==> Int y= sinx - u^3/3 = sinx - sin^3 x/ 3

Then the anti derivative of y is given by:

**Int y= sinx - (1/3) sin^3 x + C**

To detemrine the antiderivative, we'll have to calculate the indefinite integral of the given function.

We'll write the integrand (cos x)^3 = (cos x)^2*cos x.

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

We'll replace sin x by t:

sin x = t

cos x dx = dt

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

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

Int (1 - t^2)dt= t - t^3/3 + C

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

**The requested antiderivative is: F(x) = sin x - (sin x)^3/3 + C.**