`int sin^3(3x) dx` Find the indefinite integral

The integral `int sin^3(3*x) dx` has to be determined.

`int sin^3(3*x) dx`

Let `y=3x`

`dy/dx = 3`

The given integral can be written as:

`1/3int sin^3 y dy`

= `1/3int sin y * sin^2 y dy`

= `1/3int sin y * (1 - cos^2 y) dy`

Let `z = cos` y, `dz/dy = -sin y`

=> `-(1/3)*int (1 - z^2) dz `

=> `(-1/3)*(z - z^3/3)`

As `z = cosy`

=> `-1/3(cos y - (cos^3y)/3)`

And `y = 3x`

=> `-1/3(cos 3x - (cos^3(3x))/3)`

The integral `int sin^3(3*x) dx = -1/3(cos 3x - (cos^3(3x))/3) + C`

Comments

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