We have to find the integral of y = (sin x)^3

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

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

Int [ y dx] = Int [ (sin x)^3 dx]

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

let y = cos x

=> dy/dx = -sin x

=> sin x * dx = -dy

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

=> (-1 )*Int [ (1 - y^2) dy ]

=> (-1 )*(y - y^3/3)

=> y^3 / 3 - y

substitute y = cos x

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

**The required integral is (cos x)^3 / 3 - cos x + C**

The given function is y = (sin x)^3

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

We'll use Pythagorean identity:

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

We'll calculate the integral:

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

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

We'l put cos x = t.

We'll differentiate both sides:

- sin x dx = dt

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

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