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Integrating cosec...

by speeding electron
Tags: cosec, integrating
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speeding electron
#1
May24-04, 06:00 PM
P: 65
Int(cosec x)dx Let: u=cos x then: x=arcos u and: dx=-du/sqrt(1-u^2)

Int(cosec x)dx = -Int{1/sqrt(1-u^2)}^2}du = -Int {du/(1-u^2)}

= -(1/2)ln(1-u^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?
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franznietzsche
#2
May24-04, 06:30 PM
P: 1,783
Quote Quote by speeding electron
Int(cosec x)dx Let: u=cos x then: x=arcos u and: dx=du/sqrt(1-u^2)

Int(cosec x)dx = Int{1/sqrt(1-u^2)}^2}du = Int {du/(1-u^2)}

=(1/2)ln(1-u^2) + C

=(1/2)ln(sin^[2]x) +C

=ln(sin x) +C

Yet differentiating back gives cot x.
Why does this substitution not work?

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

So this makes the integral:

[tex] \int \frac{1}{u\sqrt{1-u^2}} du [/tex]

See if you can take it from there.
speeding electron
#3
May25-04, 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(1-x^2). I've edited my original post.

franznietzsche
#4
May25-04, 03:35 PM
P: 1,783
Integrating cosec...

Quote Quote by speeding electron
My query was concerning the substitution u = cos x , rather than u = sin x .

I did make a mistake, arcos x = -1/sqrt(1-x^2). I've edited my original post.

That is the wrong substitution for integrating cosec(X).[tex] cosec(x) = \frac{1}{sin(x)}[/tex].
jatin9_99
#5
May30-04, 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
arildno
#6
May30-04, 12:42 PM
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P: 12,016
Quote Quote by speeding electron
Int(cosec x)dx Let: u=cos x then: x=arcos u and: dx=-du/sqrt(1-u^2)

Int(cosec x)dx = -Int{1/sqrt(1-u^2)}^2}du = -Int {du/(1-u^2)}

= -(1/2)ln(1-u^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?
Because you make an equality out of the following non-equality:
[tex]-\int\frac{du}{1-u^{2}}\neq\frac{-1}{2}ln(1-u^{2})+C[/tex]
speeding electron
#7
May30-04, 05:23 PM
P: 65
Fine, yes, sorry about that, me being stupid again...but that integral was getting annoying.
Smin0
#8
May24-08, 10:14 AM
P: 8
Quote Quote by franznietzsche View Post
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{1-u^2}} du = dx [/tex]

So this makes the integral:

[tex] \int \frac{1}{u\sqrt{1-u^2}} du [/tex]

See if you can take it from there.
I can't :(. I've spent most of this afternoon trying this question and some other one. I know I could just use the set result, but I want to understand it, and I don't see how to integrate that last result at all :(.
Air
#9
May24-08, 11:13 AM
P: 206
There is a special method to this.

[itex]\int \mathrm{cosec} x \ \mathrm{d}x[/itex]

If you multiply this by [itex]\frac {\mathrm{cosec} x - \cot x}{\mathrm{cosec} x - \cot x}[/itex] and simplify the numerator you will get an integral of...

[itex]\int \frac{\mathrm{cosec} ^2 x - \mathrm{cosec} x \cot x}{\mathrm{cosec} x - \cot x} \ \mathrm{d}x[/itex]

Substitute [itex]u = \mathrm{cosec} x - \cot x[/itex] and it should work out beautifully. Carry on from here and post back if you still need help.
Smin0
#10
May24-08, 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{1-u^2}} du [/tex]

? Could anyone help me see how to integrate it from this?


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