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Integral of cot^4/cot^2x

User Yeritza
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2 Answers

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\int (\cot^4x)/(\cot^2x)dx = \int \cot^2x dx

using the identity:
1+\cot^2x = \csc^2x

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

=-\cot x-x +c
User Massimo Variolo
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3 votes

\bf 1+cot^2(\theta)=csc^2(\theta)\implies cot^2(\theta)=csc^2(\theta)-1\\\\ -------------------------------\\\\ \displaystyle \int~\cfrac{cot^4(x)}{cot^2(x)}\cdot dx\implies \int~\cfrac{[cot(x)]^4}{[cot(x)]^2}\cdot dx\implies \int~[cot(x)]^2\cdot dx \\\\\\ \displaystyle \int~cot^2(x)dx\implies \int~[csc^2(x)-1]dx\implies \int~csc^2(x)dx-\int1\cdot dx \\\\\\ -cot(x)-x+C
User Kupson
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