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
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