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

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

$int csc2x*dx$

= $int dx/(sin2x)$

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

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

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

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

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

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