
#1
May2404, 06:00 PM

P: 65

Int(cosec x)dx Let: u=cos x then: x=arcos u and: dx=du/sqrt(1u^2)
Int(cosec x)dx = Int{1/sqrt(1u^2)}^2}du = Int {du/(1u^2)} = (1/2)ln(1u^2) + C = (1/2)ln(sin^[2]x) +C = ln(cosec x) +C Yet differentiating back gives cot x. Why does this substitution not work? 



#2
May2404, 06:30 PM

P: 1,782

First: [tex] csc x = \frac{1}{sin x} [/tex] NOT cos x. So your substitution is wrong to begin with. If you correct that the integration should go like this: [tex] u = sin x dx [/tex] [tex] du = cos x dx [/tex] [tex] sec x du = dx [/tex] [tex] \frac{1}{\sqrt{1u^2}} du = dx [/tex] So this makes the integral: [tex] \int \frac{1}{u\sqrt{1u^2}} du [/tex] See if you can take it from there. 



#3
May2504, 03:02 PM

P: 65

My query was concerning the substitution u = cos x , rather than u = sin x .
I did make a mistake, arcos x = 1/sqrt(1x^2). I've edited my original post. 



#4
May2504, 03:35 PM

P: 1,782

Integrating cosec...That is the wrong substitution for integrating cosec(X).[tex] cosec(x) = \frac{1}{sin(x)}[/tex]. 



#5
May3004, 11:25 AM

P: 15

well your subsitution is didn't work because it is difficult to solve integration with assumption not being in the question




#6
May3004, 12:42 PM

Sci Advisor
HW Helper
PF Gold
P: 12,016

[tex]\int\frac{du}{1u^{2}}\neq\frac{1}{2}ln(1u^{2})+C[/tex] 



#7
May3004, 05:23 PM

P: 65

Fine, yes, sorry about that, me being stupid again...but that integral was getting annoying.




#8
May2408, 10:14 AM

P: 8





#9
May2408, 11:13 AM

P: 206

There is a special method to this.
[latex]\int \mathrm{cosec} x \ \mathrm{d}x[/latex] If you multiply this by [latex]\frac {\mathrm{cosec} x  \cot x}{\mathrm{cosec} x  \cot x}[/latex] and simplify the numerator you will get an integral of... [latex]\int \frac{\mathrm{cosec} ^2 x  \mathrm{cosec} x \cot x}{\mathrm{cosec} x  \cot x} \ \mathrm{d}x[/latex] Substitute [latex]u = \mathrm{cosec} x  \cot x[/latex] and it should work out beautifully. Carry on from here and post back if you still need help. 



#10
May2408, 04:43 PM

P: 8

Thanks, that worked out really well :). I was wondering though, how would one work it out from the form
[tex] \int \frac{1}{u\sqrt{1u^2}} du [/tex] ? Could anyone help me see how to integrate it from this? 


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