What is the indefinite integral of cosecant function?

[JamesGoh]

What is the indefinite integral of $cosec(\theta)$?

 

[maajdl]

http://www.wolframalpha.com/calculators/integral-calculator/

 

[HallsofIvy]

$cosc(x)= \frac{1}{sin(x)}$.

$$\int cosec(x)dx= \int \frac{1}{sin(x)}dx= \int\frac{sin(x)}{sin^2(x)}dx= \int\frac{sin(x)}{1- cos^2(x)}dx$$.

Now let u= cos(x) so that $du= -sin(x)dx$.

 

[verty]

If you know the trick to integrate $sec(x)$, try it with the co-functions instead.

 

[HakimPhilo]

Here's an alternative solution: $$\eqalign{
\int\csc x\,\mathrm dx &= \int\left(\csc x\dfrac{\csc x-\cot x}{\csc x-\cot x}\right)\mathrm dx \\
&=\int\left(\dfrac{\csc^2 x-\cot x\csc x}{\csc x-\cot x}\right)\mathrm dx.
}$$
Now use the $u$-substitution $u=\csc x-\cot x$ and you'll get: $$\int\dfrac{1}{u}\mathrm du=\ln|u|+{\cal C}=\ln|\csc x-\cot x|+{\cal C}.$$