How to integrate cot^4(x)

Note that:
∫ cot^4(x) dx = ∫ [cot^2(x) * cot^2(x)] dx
= ∫ cot^2(x)[csc^2(x) - 1] dx
= ∫ [cot^2(x)csc^2(x) - cot^2(x)] dx
= ∫ cot^2(x)csc^2(x) dx - ∫ cot^2(x) dx.

The first integral can be integrated by noting that:
d/dx cot(x) = -csc^2(x).

So:
∫ cot^2(x)csc^2(x) dx
= - ∫ cot^2(x) d[cot(x)]
= (-1/3)cot^3(x) + C.

The second integral can be integrated by re-writing cot^2(x) as csc^2(x) - 1 to get:
∫ cot^2(x) dx
= ∫ [csc^2(x) - 1] dx
= -cot(x) - x + C.

Therefore:
∫ cot^4(x) dx = ∫ cot^2(x)csc^2(x) dx - ∫ cot^2(x) dx
= (-1/3)cot^3(x) - [-cot(x) - x] + C
= (-1/3)cot^3(x) + cot(x) + x + C.

I hope this helps!

∫cot^4(x) dx

∫cot²(x)*(csc²(x) - 1) dx

∫cot²(x)*csc²(x) dx - ∫cot²(x) dx

-cot^3(x)/3 - ∫csc²(x) - 1 dx

-cot^3(x)/3 + cot(x) + x + C