How do you find the integral of #csc^2x/cot^3x dx#?

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Answer 1 · 2018-05-22T09:40:03

#int csc^2x/cot^3x# #"d"x=1/2tan^2x+"c"#

We want to find #intcsc^2x/cot^3x# #"d"x#. Note that

#csc^2x/cot^3x=1/sin^2xsin^3x/cos^3x=sinx/cos^3x=sec^2xtanx#

So,

#intcsc^2x/cot^3x# #"d"x=intsec^2xtanx# #"d"x#

We now perform the natural substitution #u=tanx# and #"d"u=sec^2x# #"d"x#

Hence,

#intsec^2xtanx# #"d"x=intu# #"d"u=1/2u^2=1/2tan^2x+"c"#