# Math Help - integral

1. ## integral

how do i integrate:
$\int{cot^2(x)} dx$

2. Originally Posted by scorpion007
how do i integrate:
$\int{cot^2(x)} dx$
Hi,

$\int{cot^2(x)} dx$ = $\int{\frac{cos^2(x)}{sin^2(x)}} dx$

$\int{\left(\frac{cos^2(x)}{sin^2(x)}+1-1 \right)} dx$

$\int{\left(\frac{cos^2(x)}{sin^2(x)}+\frac{sin^2(x )}{sin^2(x)}-1 \right)} dx$

$\int{\left( \frac{cos^2(x)+sin^2(x)}{sin^2(x)}-1 \right)} dx$

$\int{\left( \frac{1}{sin^2(x)}-1\right)} dx$=-cot(x)-x+C

EB

3. Originally Posted by earboth
Hi,

$\int{cot^2(x)} dx$ = $\int{\frac{cos^2(x)}{sin^2(x)}} dx$
From here you are going around the houses to get where you are going:

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

$\int{\left(\frac{cos^2(x)}{sin^2(x)}+1-1 \right)} dx$

$\int{\left(\frac{cos^2(x)}{sin^2(x)}+\frac{sin^2(x )}{sin^2(x)}-1 \right)} dx$

$\int{\left( \frac{cos^2(x)+sin^2(x)}{sin^2(x)}-1 \right)} dx$

$\int{\left( \frac{1}{sin^2(x)}-1\right)} dx$=-cot(x)-x+C

EB
But you are using the standard integral:

$\int \csc^2(x) dx =-\cot(x)$,

in which case you just could have used the sandard integral:

$\int{\cot^2(x)} dx =-\cot(x)-x$.

RonL

4. Originally Posted by CaptainBlack
From here you are going around the houses to get where you are going:...

RonL
It's so much fun to go the long way round...

Actually

- the functions sec, csc, cot were not taught in German schools because you can express them by the standard trig functions;

- I was very proud about me that I detected the solution I wrote to scorpion;

- you are right.

EB

5. Hello, scorpion007!

$\int \cot^2(x)\,dx$

Use the identity: . $\cot^2\theta\:=\:\csc^2\theta - 1$

$\int\cot^2x\,dx\;=\;\int\left(\csc^2x - 1\right)\,dx\;=\;-\cot x - x + C$

Edit: corrected the typo.
. . . .Thanks for the heads-up, Scorpion!

6. thanks very much guys.