Respuesta :
[tex]\bf 1+cot^2(\theta)=csc^2(\theta)\implies cot^2(\theta)=csc^2(\theta)-1\\\\
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\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
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\displaystyle \int~cot^2(x)dx\implies \int~[csc^2(x)-1]dx\implies \int~csc^2(x)dx-\int1\cdot dx
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-cot(x)-x+C[/tex]
[tex]\int \frac{\cot^4x}{\cot^2x}dx = \int \cot^2x dx [/tex]
using the identity: [tex]1+\cot^2x = \csc^2x[/tex]
[tex]\int \cot^2x dx = \int( \csc^2x - 1 )dx[/tex]
[tex]=-\cot x-x +c[/tex]
using the identity: [tex]1+\cot^2x = \csc^2x[/tex]
[tex]\int \cot^2x dx = \int( \csc^2x - 1 )dx[/tex]
[tex]=-\cot x-x +c[/tex]
