What's the integral of #int (cscx)^2 dx#?

1 Answer
Nov 2, 2015

#int (cscx)^2 dx = -cotx+C#

Explanation:

The derivative of #cotx# is #csc^2x#,

so the integral of #csc^2x# is #cotx + C#

If you really want the integral of the integral, then use:

#int[int (cscx)^2 dx] dx = int [-cotx+C] dx# where #C# is an arbitrary constant

# = int (-cosx/sinx + C) dx# where #C# is an arbitrary constant

# = -lnabssinx + Cx +D# where #C,D# are arbitrary constants

# = lnabscscx + Cx +D# where #C,D# are arbitrary constants