**[tex]\int csc dx = ?[/tex]**

**I just am seeking confirmation as to whether or not I did this correctly. I apologize beforehand for any formatting errors.**

[tex]\int csc(x) dx = [/tex] [tex]\int 1/sin(x) dx[/tex]

u = sin(x)

du = cos(x)dx

du / cos(x) = dx

u

u

[tex]\sqrt{}[/tex]{u^2-1} = cos(x)

du / [tex]\sqrt {u

Therefore

[tex]\int du / u\sqrt{}[/tex]{u

And, since I cannot remember the arc property, this is as for as I was able to get.

[tex]\int csc(x) dx = [/tex] [tex]\int 1/sin(x) dx[/tex]

u = sin(x)

du = cos(x)dx

du / cos(x) = dx

u

^{2}= sin^{2}(x)u

^{2}= 1 - cos^{2}(x)[tex]\sqrt{}[/tex]{u^2-1} = cos(x)

du / [tex]\sqrt {u

^{2}-1} = dx[/tex]Therefore

[tex]\int du / u\sqrt{}[/tex]{u

^{2}-1}And, since I cannot remember the arc property, this is as for as I was able to get.

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