Bonus Homework Problem: Integral of csc

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[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

u2 = sin2(x)
u2 = 1 - cos2(x)
[tex]\sqrt{}[/tex]{u^2-1} = cos(x)
du / [tex]\sqrt {u2-1} = dx[/tex]

Therefore

[tex]\int du / u\sqrt{}[/tex]{u2-1}

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

Homework Helper
3,305
2
try u = ln(tan(x/2))
 
31,930
3,893
[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

u2 = sin2(x)
u2 = 1 - cos2(x)

This is an interesting approach, but you have a problem in the next line. The intermediate step is: u2 - 1 = - cos2(x).
The right side is always <= 0, so taking its square root almost always gives you imaginary numbers.

[tex]\sqrt{}[/tex]{u2-1} = cos(x)
du / [tex]\sqrt {u2-1} = dx[/tex]

Therefore

[tex]\int du / u\sqrt{}[/tex]{u2-1}

And, since I cannot remember the arc property, this is as for as I was able to get.
That's an interesting approach, but there's a problem here:
 

HallsofIvy

Science Advisor
41,626
821
I would integrate [itex]\int dx/cos(x)[/itex] by multiplying both numerator and denominator by cos(x): [itex]\int cos(x)dx/cos^2(x)= \int cos(x)dx/(1- sin^2(x))[/itex] and then let u= sin(x).
 

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