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
From here you are going around the houses to get where you are going:
$\displaystyle \int{\frac{\cos^2(x)}{\sin^2(x)}} dx= \int{\frac{1-\sin^2(x)}{\sin^2(x)}} dx=\int{\frac{1}{\sin^2(x)}-1} dx$
But you are using the standard integral:$\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
$\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
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
Hello, scorpion007!
$\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!