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Posted January 21st, 2010 by alpha
in

$ \int cot^4x dx $

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answer

On January 21st, 2010 Structure says:

$ \int cot^4x dx $=$ \int (\cot^4x+\cot^2x-\cot^2x-1+1) dx =-\frac{1}{3}\cot^3x+\cot x+x+\Mathcal{C} $

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