You should remember that `cscx = 1/ sin x` , hence `csc 3x= 1/(sin 3x)` such that:

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

You may use the alternate form of fundamental formula of trigonometry such that:

`1/(sin^2 3x) = cot^2 3x + 1`

`int 1/((sin 3x)^4) dx = int (cot^2 3x + 1)1/(sin^2...

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You should remember that `cscx = 1/ sin x` , hence `csc 3x= 1/(sin 3x)` such that:

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

You may use the alternate form of fundamental formula of trigonometry such that:

`1/(sin^2 3x) = cot^2 3x + 1`

`int 1/((sin 3x)^4) dx = int (cot^2 3x + 1)1/(sin^2 3x) dx`

You need to substitute t for `3x` such that:

`3x = t => 3dx = dt => dx = (dt)/3`

Changing the variable yields:

`int (cot^2 3x + 1)1/(sin^2 3x) dx = (1/3)int (cot^2t + 1)1/(sin^2 t) dt`

You should come up with the following substitution such that:

`cot t = u => -1/(sin^2 t)dt = du`

Changing the variable yields:

`(1/3)int (cot^2 t + 1)1/(sin^2 t) dt = -(1/3)int (u^2 + 1) du`

Using the property of linearity yields:

`-(1/3)int (u^2 + 1) du = -(1/3)(int u^2 du + int du)`

`-(1/3)int (u^2 + 1) du = -(1/3)(u^3/3 + u) + c`

Substituting back `cot t` for u yields:

`(1/3)int (cot^2 t + 1)1/(sin^2 t) dt = -(1/3)((cot^3 t)/3 + cot t) + c`

Substituting back `3x` for t yields:

`int csc^4 (3x)dx = -(1/3)((cot^3 (3x))/3 + cot (3x)) + c`

**Hence, evaluating the given integral yields `int csc^4 (3x)dx = -(1/3)((cot^3 (3x))/3 + cot (3x)) + c.` **