Integrating csc(x): Easier Than You Think

[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 :/

 

[marlon]

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

marlon

 

[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...

 

[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

 

[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})}


Can you please clarify?

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

 

[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?)

 

[DrKareem]

yes, nice method :)

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

 

[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.

 

[uart]

Still i would like someone to explain the question of my last post.

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.

 

[DrKareem]

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

Thanks a lot for your help guys :)