# integral

• Oct 25th 2006, 08:58 PM
scorpion007
integral
how do i integrate:
$\displaystyle \int{cot^2(x)} dx$
• Oct 26th 2006, 04:39 AM
earboth
Quote:

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

Hi,

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

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

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

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

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

EB
• Oct 26th 2006, 05:03 AM
CaptainBlack
Quote:

Originally Posted by earboth
Hi,

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

From here you are going around the houses to get where you are going:

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

Quote:

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

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

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

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

EB
But you are using the standard integral:

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

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

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

RonL
• Oct 26th 2006, 05:50 AM
earboth
Quote:

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
• Oct 26th 2006, 08:27 AM
Soroban
Hello, scorpion007!

Quote:

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

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

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

Edit: corrected the typo.
. . . .Thanks for the heads-up, Scorpion!
• Oct 26th 2006, 07:24 PM
scorpion007
thanks very much guys.