How to integrate #xcsc^2x#?

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Answer 1 · 2018-04-01T16:13:38

#int x csc^2x dx =x cotx - ln abs sinx +C#

Considering that:

#d/dx cotx = csc^2x#

we can integrate by parts:

#int x csc^2x dx = int x d(cotx)#

#int x csc^2x dx =x cotx - int cotx dx#

#int x csc^2x dx =x cotx - int cosx/sinx dx#

#int x csc^2x dx =x cotx - int (d (sinx))/sinx #

#int x csc^2x dx =x cotx - ln abs sinx +C#