# How to find out the indefinite integral of the cube of sinx?

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`int sin^3(x)dx`

using the identity `sin^3x =-1/4sin(3x) +3/4 sin(x)`

we get

`int sin^3(x)dx = int (-1/4 sin(3x) + 3/4 sin(x))dx=

-1/4(-1/3 cos(3x)) +3/4 (-cos(x)) + C`

`int sin^3(x)dx=1/12cos(3x)-3/4cos(x)+C`

To reconcile with the above calculation

`cos3x=4cos^3x-3cosx` so

`1/12cos3x-3/4cosx=1/12(4cos^3x-3cosx)-3/4cosx`

`=1/3cos^3x-1/4cosx-3/4cosx=1/3cos^3x-cosx`

`int sin^3x=1/12cos(3x)-3/4cosx+C=1/3cos^3x-cosx+C```

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

According to Pythagorean identity, we'll get:

(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 evaluate `int` (cosx)^2*sin xdx using substitution:

cos x = t

We'll differentiate both sides:

-sin x dx = 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**