Integrating cosec

1. May 24, 2004

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?

Last edited: May 25, 2004
2. May 24, 2004

franznietzsche

First:

$$csc x = \frac{1}{sin x}$$

NOT cos x. So your substitution is wrong to begin with. If you correct that the integration should go like this:

$$u = sin x dx$$

$$du = cos x dx$$

$$sec x du = dx$$

$$\frac{1}{\sqrt{1-u^2}} du = dx$$

So this makes the integral:

$$\int \frac{1}{u\sqrt{1-u^2}} du$$

See if you can take it from there.

3. May 25, 2004

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.

Last edited: May 25, 2004
4. May 25, 2004

franznietzsche

That is the wrong substitution for integrating cosec(X).$$cosec(x) = \frac{1}{sin(x)}$$.

5. May 30, 2004

jatin9_99

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

6. May 30, 2004

arildno

Because you make an equality out of the following non-equality:
$$-\int\frac{du}{1-u^{2}}\neq\frac{-1}{2}ln(1-u^{2})+C$$

7. May 30, 2004

speeding electron

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

8. May 24, 2008

Smin0

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 :(.

9. May 24, 2008

Air

There is a special method to this.

$\int \mathrm{cosec} x \ \mathrm{d}x$

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

$\int \frac{\mathrm{cosec} ^2 x - \mathrm{cosec} x \cot x}{\mathrm{cosec} x - \cot x} \ \mathrm{d}x$

Substitute $u = \mathrm{cosec} x - \cot x$ and it should work out beautifully. Carry on from here and post back if you still need help.

10. May 24, 2008

Smin0

Thanks :)!

Thanks, that worked out really well :). I was wondering though, how would one work it out from the form

$$\int \frac{1}{u\sqrt{1-u^2}} du$$

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