We have:

`6intcot^3x\ dx =6int(cot^2 x)cot\ x\ dx`

`=6int(csc^2x-1)cot\ x\ dx`

`=6(intcsc^2 x\ cot\ x\ dx` `{:-intcot\ x\ dx)`

Take the `intcsc^2x\ cot\ x\ dx` integral first:

Let `u = cot\ x`, then `du = -csc^2x\ dx`

So

`intcsc^2x\ cot\ x\ dx =-intu\ du`

`=-u^2/2+C`

`=-(cot^2x)/2+C`

For the second integral, from our table, we have

`intcot\ x\ dx=ln\ |sin\ x|+C`

Returning to the main question:

`6intcot^3x\ dx =6(intcsc^2x\ cot\ x\ dx-` `{:intcot\ x\ dx)`

`=6(-(cot^2x)/2-ln\ |sin\ x|)+K`

`=-3\ cot^2x-6\ ln\ |sin\ x|+K`

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