1. ## Trig Integration

Ok I am not sure if I am getting this right or not. My answer is slightly different to an answer I am getting from a symbolic calculator.

$\int \cot{x} \csc^{2}{x}dx$

So I substituted

$u=\cot{x}$

and

$du=-\csc^{2}xdx$

or

$-du=csc^{2}xdx$

so

$\int \cot{x} \csc^{2}xdx = \int-udu$
$=\frac{-u^{2}}{2}+C$
$=\frac{-cot^{2}x}{2}+C$

But the answer the calculator gives me is

$\frac{-csc^{2}x}{2}+C$

Can someone point out where I might have gone wrong. Thanks!

3. ## Re: Trig Integration

Amended! Thanks!

4. ## Re: Trig Integration

You can always check this by calculating the derivative of the primitive function you found:
$\frac{d}{dx}\left[\frac{-1}{2}\cot^2(x)+C\right]$
$=\frac{-1}{2}\cdot \left(\frac{d}{dx}\cot^2(x)\right)$
$=\frac{-1}{2}\cdot \left(2\cdot \cot(x)\cdot \frac{-1}{\sin^2(x)}\right)$
$=\cot(x)\cdot \frac{1}{\sin^2(x)}=\cot(x)\cdot \csc^2(x)$
= the integrand

5. ## Re: Trig Integration

you can arrange the integral
$\int\csc(x)\csc(x)\cot(x)$ dx and substitute $\csc(x)=u$

6. ## Re: Trig Integration

Originally Posted by Siron
You can always check this by calculating the derivative of the primitive function you found:
$\frac{d}{dx}\left[\frac{-1}{2}\cot^2(x)+C\right]$
$=\frac{-1}{2}\cdot \left(\frac{d}{dx}\cot^2(x)\right)$
$=\frac{-1}{2}\cdot \left(2\cdot \cot(x)\cdot \frac{-1}{\sin^2(x)}\right)$
$=\cot(x)\cdot \frac{1}{\sin^2(x)}=\cot(x)\cdot \csc^2(x)$
= the integrand
Ah yes I should have thought of that, it was quite late at night when I was doing this tutorial.

7. ## Re: Trig Integration

Wouldn't that give

$-\frac{\csc^{2}x}{2}+C$

$\int\csc(x)\csc(x)\cot(x)$ dx and substitute $\csc(x)=u$