Integral of Cot^9(x)Csc^2(x)dx

The derivative of cot(x) is -csc^2(x)

Let u = cot(x), then du = -csc^2(x)dx and -du = csc^2(x)dx

So we now have ∫ -u^9 du = -((u^10) / 10) + C

Substituting cot(x) back for u we get the following answer: ((-cot^(10)) / 10) + C

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