How do you integrate #int (csc2x)dx#?

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Answer 1 · 2018-06-05T10:41:27

#I=1/2ln|csc2x-cot2x|+c#

We know that,

#color(red)(intcsctheta d theta=ln|csctheta- cottheta|+c)..........(1)#

Here,

#I=int(csc2x)dx#

Let, #2x=u=>2dx=du=>dx=1/2du#

#I=intcscu*1/2du#

#=1/2intcscudu...tocolor(red)(Apply(1)#

#=1/2ln|cscu-cotu|+c ,where, u=2x#

#=1/2ln|csc2x-cot2x|+c#

Answer 2 · 2018-06-05T13:58:30

#int csc2x*dx=1/2ln(tanx)+C#

#int csc2x*dx#

=#int dx/(sin2x)#

=#int dx/(2sinx*cosx)#
=#int ([(sinx)^2+(cosx)^2]*dx)/(2sinx*cosx)#

=#1/2int (sinx*dx)/cosx+1/2int (cosx*dx)/sinx#

=#-1/2ln(cosx)+1/2ln(sinx)+C#

=#1/2ln(sinx/cosx)+C#

=#1/2ln(tanx)+C#