# What is the antiderivative of the function f(x) = sin^3 x?

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We'll calculate the antiderivative F(x) integrating the given function.

F(x) is a function such as dF/dx = f(x).

Int f(x)dx = F(x) + C

We'll write the function as a product:

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

We'll integrate both sides:

Int (sinx)^3dx = Int [(sinx)^2*sin x]dx

We'll write (sinx)^2 = 1 - (cosx)^2

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

We'll remove the brackets:

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

We'll solve Int (cosx)^2*sin xdx using substitution technique:

cos x = t

We'll differentiate both sides:

cos xdx = dt

We'll re-write the integral, changing the variable:

Int (cosx)^2*sin xdx = Int t^2dt

Int t^2dt = t^3/3 + C

Int (cosx)^2*sin xdx = (cos x)^3/3 + C

Int (sinx)^3dx = Int sin xdx - Int (cosx)^2*sin xdx

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

f(x) = sin^3x.

To find the anti derivative of f(x). Or we have to find Integral sin^3x dx.

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

f(x) = (1-cos^x) sinx.

Therfore integral sin^3 x = Integral (1-cos^2) sinx.

Let cos x = t. Then d/dx cosx = sinx.

d/dx t = -sinx. So sinx dx = -dt.

Therefore Integral (1-cos^2x) dx = Integral (1-t^2) (-dt).

Integral (1-cos^2) dx = Integral (t^2-1) dt = t^3/3 -t.

Therefore Integral sin^3x = (cosx)^3/3 - cosx +C.

Therefore the antiderivative of sin^3x = (cosx)^3 /3 - cosx +C.