# Integrating csc(x)

1. Nov 21, 2004

### DrKareem

haven't found a way of doing it so far. I have a feeling that it's extremely easy, and i'm missing how to do it somehow :/

Last edited: Nov 21, 2004
2. Nov 21, 2004

### marlon

do you mean cosec(x), the cosecans ??? If so, just use the t=tan(x/2) formula's...

marlon

3. Nov 21, 2004

### DrKareem

I'm not sure what formula you're talking about.

$$\int \csc(x) dx$$

If you take $$t=tan(\frac{x}{2})$$,

you'd get:

$$\frac {d}{dx} \tan(x)= \frac{1}{2}.sec^2(x)$$

Not sure how to go from there...

Last edited: Nov 21, 2004
4. Nov 21, 2004

### arildno

What marlon meant, is the following:
$$csc(x)=\frac{1}{\sin(x)}=\frac{\cos^{2}(\frac{x}{2})+\sin^{2}(\frac{x}{2})}{2\sin(\frac{x}{2})\cos(\frac{x}{2})}=\frac{1+tan^{2}(\frac{x}{2})}{2tan(\frac{x}{2})}$$

Substitute $$u=tan(\frac{x}{2})$$
This implies:
$$\frac{du}{dx}=\frac{1}{2}\frac{1}{\cos^{2}(\frac{x}{2})}=\frac{1}{2}(u^{2}+1)$$
Or:
$$dx=\frac{2du}{u^{2}+1}$$
Hence, we have:
$$\int{csc(x)}dx=\int\frac{du}{u}=ln|u|+C=ln|tan(\frac{x}{2})|+C$$

5. Nov 22, 2004

### DrKareem

I'm not sure how you did this equality:

$$\frac{\cos^{2}(\frac{x}{2 })+\sin^{2}(\frac{x}{2})}{2\sin(\frac{x}{2})\cos(\ frac{x}{2})}=\frac{1+tan^{2}(\frac{x}{2})}{2tan(\frac{x}{2})}$$

Other than that, it's all clear, thank you very much.

Last edited: Nov 22, 2004
6. Nov 22, 2004

### happenstantially

$\int \csc x = \int \csc x \left(\frac{\csc x - \cot x}{\csc x - \cot x}\right) = \int \frac{du}{u} = \ln |csc x - cot x|$
(is that what we're talking about?)

Last edited: Nov 22, 2004
7. Nov 22, 2004

### DrKareem

yes, nice method :)

Still i would like someone to explain the question of my last post.
Thank you :)

8. Nov 22, 2004

### happenstantially

separate the fractions and simplify
$\frac{\cos u}{2\sin u}+ \frac{\sin u}{2\cos u} =$
$\frac{1}{2\tan u} + \frac{tan u}{2}$
get a common denominator and you're done.

9. Nov 22, 2004

### uart

Just divide by cos^2(x/2) in both the numerator and the denominator of the LHS of the expression in question and it will drop straight out.

10. Nov 22, 2004

### DrKareem

Yes, excellent, so know i know i am stupid hehe :)

Thanks alot for your help guys :)